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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Statement-1 `sum_(r=0)^(n) r ""^(n)C_(r) x^(r) (-1)^(r) = nx (1 - x)^(n -1)` Statement-2:` sum_(r=0)^(n)r ""^(n)C_(r) x^(r) (-1)^(r) =0`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - d We have, `sum_(r=0)^(n) r ""^(n)C_(r) x^(r) (-1)^(r) = sum_(r=0)^(n) (-r)(n)/(r) ""^(n-1)C_(r-1) x^(r-1)x(-1)^(r-1)` `rArr sum_(r=0)^(n) r ""^(n)C_(r) x^(r) (-1)^(r) = -nx sum_(r=1)^(n) ""^(n-1)C_(r-1) x^(r-1)x(-1)^(r-1)` `rArr sum_(r=0)^(n) r ""^(n)C_(r) x^(r) (-1)^(r) = -nx (1 -x)^(n-1)` ...(i) So, statement-1 is not true. Replacing x by 1 in (i) , we get ` sum_(r=0)^(n) r ""^(n)C_(r)(-1)^(r) =0` So, statement-2 is true. |
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| 52. |
Statement-1 : ` sum_(r=0)^(n) r^(2) ""^(n)C_(r) x^(r) = n (n-1) x^(2) (1 + x)^(n-2) + nx (1 +x)^(n-1)`Statement-2: `sum_(r=0)^(n) r^(2) ""^(n)C_(r) = n (n-1)2^(n-2)+ n2^(n-1)` .A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - b We have, ` sum_(r=0)^(n) r^(2) ""^(n)C_(r) x^(r)` ` = sum_(r=0)^(n) {r( r-1)+r}""^(n)C_(r)x^(r)` `= sum_(r=0)^(n) r(r -1) ""^(n)C_(r) x^(r) sum_(r=0)^(n) r ""^(n)C_(r) x^(r)` `sum_(r=0)^(n) r(r -1) xx(n)/(r)xx(n-1)/(r-1) ""^(n-2)C_(r-2) x^(r) -2x^(2) + sum_(r=1)^(n)r xx (n)/(r) xx""^(n-1)C_(r-1) x^(r-1) xxx` ` = n (n-1) x^(2) sum_(r=2)^(n) ""^(n-1)C_(r-2) x^(r-2) + nx sum_(r=1)^(n) ""^(n-1)C_(r-1) x^(r-1)` `= n(n-1) x^(2) (1 + x)^(n-2) +nx (1 + x)^(n-1)` So, statement-2 s also true. Clearly, statement-1 is a correct expanation for statememt -2. But, statement-2 is not a correct expanation for statement-1. |
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| 53. |
Find the greatest value of the term independent of `x`in the expansion of `(xsinalpha+(cosalpha)/x)^(10)`, where `alpha in Rdot`A. `2^(5)`B. `(10!)/((5!)^(2))`C. `(10!)/(2^(5) xx(5 !))^(2)`D. none of these |
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Answer» Correct Answer - c Let `(r + 1)^(th)` term be indepandent of x. We have, `T_(r + 1) = ""^(10)C_(r) (x sin alpha)^(10 - r) ((cos alpha)/(x))^(r)` `rArr T_(r +1) = ""^(10)C_(r) x ^(10 - 2r) (sin alpha)^(10 - r) (cos alpha )^(r)` If it is independent of x, then r = 5 `therefore ` Term independent of x, `= T_(6)= ""^(10)C_(5)(sin alpha cos alpha)^(5) = ""^(10)C_(5) xx2 ^(-5) (sin 2 alpha)^(5)` Clearly , it is greatest when 2 `alpha = pi 2` and its greatest value is `""^(10)C_(5) xx2^(-5) = (10!)/(2^(5) (5!)^(2))` |
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| 54. |
Find the term independent of `x`in the expansion of `(1+x+2x^3)[(3x^2//2)-(1//3)]^9`A. `1//3`B. `19//54`C. `17//54`D. `1//4` |
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Answer» Correct Answer - c We have, `(1 + x + 2x^(3))((3)/(2) x^(2)-(1)/(3x))^(9)` `(1 + x + 2x^(3)){sum_(r=0)^(9)""^(9)C_(r) ((3)/(2) x^(2))^(9-r)(-(1)/(3x))^(r)}` `(1 + x + 2x^(3)){sum_(r=0)^(9)""^(9)C_(r) ((3)/(2))^(9-r)(-(1)/(3))^(r)x^(18 - 3r)}` `= {sum_(r=0)^(9)""^(9)C_(r) ((3)/(2))^(9-r)(-(1)/(3))^(r)x^(18 - 3r)}` `+ {sum_(r=0)^(9)""^(9)C_(r) ((3)/(2))^(9-r)(-(1)/(3))^(r)x^(19 - 3r)}` `+2 {sum_(r=0)^(9)""^(9)C_(r) ((3)/(2))^(9-r)(-(1)/(3))^(r)x^(21 - 3r)}` Clearly, first and third expansions contain term idepen- dent of x and obtained by equation 18 - 3r = 0 and 21 -3r = 0 respectivley. So, coefficient of the term independent of x `""^(9)C_(6) ((3)/(2))^(9-6)(1-(1)/(3))^(6) +2{""^(9)C_(7)((3)/(2))^(9-7) (-(1)/(3))^(7)}` ` (7)/(18) - (2)/(27) = (17)/(54)` . |
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| 55. |
The coefficient of the term independent of x in the expansion of `((x+1) /(x^(2//3) - x^(1//3) + 1 )- (x -1)/(x - x ^(1//2)))^(10)`A. 210B. 105C. 70D. 112 |
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Answer» Correct Answer - a We have, = `(x+1) /(x^(2//3) - x^(1//3) + 1 )- (x -1)/(x - x ^(1//2))` ltbrge `((x^(1//3))^(3) + 1^(3)) /(x^(2//3) - x^(1//3) + 1 )- (x -1)/( x ^(1//2)-1)` = `((x^(1//3)+ 1) x^(2//3) - x^(1//3) + 1 )/(x^(2//3) - x^(1//3) + 1 )- (x^(1//2) +1)/( x ^(1//2))` ` x^(1//3) + 1 + 1 - x ^(-1//2) = x^(1//3) - x^(-1//2)` `((x+1)/(x^(2//3) - x^(1//3) + 1 )-(x-1)/ (x-x^(1//2)))^(10) = (x^(1//3) - x ^(-1//2))^(10)` Let `T_(r +1)` be the general term in `(x^(1//3) - x ^(-1//2))^(10)`. Then, `t_(r +1) = ""^(10)C_(r) (x^(1//3)^(10-r)(-1)^(r) (x ^(-1//2))^(r)` For this term sto be independent of x, we must have `(10 - r)/(3) - (r)/(2) = 0 rArr 20 - 2r - 3r = 0 rArr r = 4` So, required coefficient = `""^(10)C_(4) (-1)^(4) = 210` |
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| 56. |
Consider the expansion `(x^(2)+(1)/(x))^(15)`. What is the independent term in the given expansion ?A. `""^(15)C_(9)`B. 0C. `-""^(15)C_(9)`D. 1 |
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Answer» Correct Answer - c Let `(r +1)^(th) ` term be the constant term in the expansion of `(x^(3) - (1)/(x^(2)))^(15)` . `because T_(r+1) = ""^(15)C_(r) (x^(3))^(15-r) (-(1)/(x^(2)))^(r)` is independent of x `rArr T_(r +1) = ""^(15)C_(r) x^(45 - 5r) (-1)^(r)` is independent of x `rArr 45 - 5r = 0 rArr r = 9` Thus , tenth is independent of x and is given by `T_(10) = ""^(15)C_(9) (-1)^(9) = - ""^(15)C_(9)` |
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| 57. |
Statement-1: ` sum_(r =0)^(n) (r +1)""^(n)C_(r) = (n +2) 2^(n-1)` Statement -2: ` sum_(r =0)^(n) (r+1) ""^(n)C_(r) x^(r) = (1 + x)^(n) + nx (1 + x)^(n-1)`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A We have , ` sum_(r =0)^(n) (r+1) ""^(n)C_(r) x^(r) = sum_(r =0)^(n) r. ""^(n)C_(r) x^(r)+ sum_(r =0)^(n) r. ""^(n)C_(r) x^(r)` `rArr sum_(r =0)^(n) (r+1) ""^(n)C_(r) x^(r) = sum_(r =1 )^(n) rxx(n)/(r)xx ""^(n-1)C_(r-1) x^(r-1)xxx+ sum_(r =0)^(n) ""^(n)C_(r) x^(r)` `rArr sum_(r =0)^(n) (r+1) ""^(n)C_(r) x^(r) =nx sum_(r =1)^(n) ""^(n-1)C_(r-1) x^(r-1)xxx+ sum_(r =0)^(n) ""^(n)C_(r) x^(r)` `rArr sum_(r=0)^(n) (r+1)""^(n)C_(r) x^(r) = nx (1 + x)^(n-1) + 2 ^(n) (n +2)2^(n-1)` So, statement -1 is also ture ans statement-2 is a correct expanation for statement-1 . |
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| 58. |
The term independent of x in the expansion of `((1)/(x^(2)) + (1)/(x) +1 + x + x^(2))^(5)`, isA. 381B. 441C. 439D. 359 |
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Answer» Correct Answer - a We have, `((1)/(x^(2)) + (1)/(x) + 1+ x + x^(2))^(5) = {(1)/(x^(2))((1 -x^(5))/(1 - x))}^(5) = (1)/(x^(10)) (1 - x^(5))^(5)(1 - x)^(-5)` Clearly, the term independent of x in ` ((1)/(x^(2)) + (1)/(x) + 1 + x + x^(2))^(5)` is equal to the coefficient of `x^(10)` in `(1 - x ^(5))^(5) (1 - x)^(-5)` . Now, `(1 - x^(5))^(5) (1 - x )^(-5) = (""^(5)C_(0) - ""^(5)C_(1)x^(5)+""^(5)C_(2)x^10-...)(sum_(r=0)^(infty)""^(4 +r)C_(r)x^(r))` `- (""^(5)C_(0) - ""^(5)C_(1)x^(5)+""^(5)C_(2)x^10-...)(sum_(r=0)^(infty)""^(4 +r)C_(4)x^(r))` `therefore ` Coefficient of `x^(10)` in `(1 - x ^(5))^(5) (1 - x)^(-5)` `=""^(5)C_(0) xx ""^(14)C_(4)-""^(5)C_(1)xx""^(9)C_(4)+""^(5)C_(2)xx""^(4)C_(4)= 381` |
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| 59. |
Stetemet - 1: `sum_(r=0)^(n) r. ""^(n)C_(r) = n 2^(n-1)` Statement-2: ` sum_(r=0)^(n) r. ""^(n)C_(r) x^(r) = n (1 + x )^(n-1) x`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - a We have, `sum_(r=0)^(n) r. ""^(n)C_(r) x^(r) = sum_(r=0)^(n) r (n)/(r) ""^(n-1)C_(r-1) x^(r-1) xxx` `rArr sum_(r =0)^(n) r ""^(n)C_(r) x^(r) = nx sum_(r=1)^(n) ""^(n-1)C_(r-1) x^(r -1)` `rArr sum_(r=0)^(n) r ""^(n)C_(r) x^(r) = nx (1 + x)^(n-1)` ...(i) So, statement-2 is true. Replacing x by in (i), we get `sum_(r=0)^(n) r ""^(n)C_(r) = n2^(n-1)` So, statement -1 is also true stetement -2 is a correct expansion for statement-1. |
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| 60. |
The value of `sum_(r=0)^(n) r(n -r) (""^(n)C_(r))^(2)` is equal toA. `n^(2) ""^(2n -1)C_(n -1)`B. `n^(2) """^(2n-2)C_(n)`C. `n^(2) ""^(2n)C_(n -1)`D. `n^(2) ""^(2n -1)C_(n)` |
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Answer» Correct Answer - b `sum_(r=0)^(n) (n-r)(""^(n)C_(r))^(2)` `= n sun_(r=0)^(n) (r ""^(n)C_(r)) (""^(n)C_(r)) - sum_(r=1)^(n) (r ""^(n)C_(r))^(2)` `= n^(2) (sum_(r=1)^(n) ""^(n-1)C_(r-1) ""^(n)C_(r)) - n^(2) {sum_(r=1)^(n) (""^(n-1)C_(r-1))^(2)}` `= n^(2)` { Coeff. of` "" x^(n-1)` in` (1 + x )^(2n-1)}` ` - n^(2) `{Coeff. of `x^(n-1)` in`(1 + x)^(2n-2)} ` `= n^(2) ""^(2n-1)c_(n-1) - n^(2) ""^(2n -2)C_(n-1)` `n^(2) {((2n -1)1)/(n!(n-1)!) - ((2n -2)!)/((n-1)!(n-1)!) -n^(2) ((2n -2)!)/((n-2)!n)` ` n^(2) ""^(2n-2)C_(n)` |
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| 61. |
The sum of rational term(s) in `(sqrt3 + 2^(1/3) + 5^(1/4))^8` is equal toA. 3150B. 336C. 3486D. 3592 |
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Answer» Correct Answer - d The general term in the expansion of `(sqrt(3) + root(3)(2)+ root(4)(5))^(8)` is `(8!)/(r!s!t!)(3)^(r//2) (2)^(s//3)(5)^(1//4)`, where ` r + s + t = 8` For rational terms, we must have, `(i) r = 3 , s=6, t=0 " "(b) r = 4 , s = 0, t= 4` `(c) r = 0 , s=0, t=8 " "(d) r = 8 , s = 0, t= 0` `therefore ` Sum of rational terms `(8!)/(3!6!0!)xx3xx2^(2)xx5^(0) + (8!)/(4!0!4!)xx3^(2)xx2^(0)xx5` `(8!)/(0!0!8!)xx3^(0)xx2^(0)xx5^(2) + (8!)/(8!0!0!)xx3^(4)xx2^(0)xx5^(0)` = 3592 |
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| 62. |
The number of real negative terms in the binomial expansion of `(1+i x)^(4n-2),n in N ,x >0`is`n`b. `n+1`c. `n-1`d. `2n`A. nB. n+1C. n-1D. 2n |
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Answer» Correct Answer - b Let `T_(r+1)` be the `(r +1)^(th)` term in the binomial expanion of `(1 + ix)^(4n -2)` . Then `T_(r+1) = ""^(4n-2)C_(r)(ix)^(r) = ""^(2n-2)C_(r)x^(r)i^(r)` Clearly, `T_(r +1) ` is real negative , if ` r=2 ,6,10,14,…` These values of r from an A.P. with term 2 and common difference 4. lies between 2 and 4n -2 `therefore 2 le 4k - 2 le 4n - 2 rArr 1 le k le n` Hence, there are n real negative terms. |
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| 63. |
If `(3+asqrt2)^100+(3+bsqrt2)^100=7+5sqrt2` number of pairs (a, b) for which the equation is true is, (a, b are rational numbers)A. 1B. 6C. 0D. infinite |
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Answer» Correct Answer - C We have, `(3 + asqrt(2))^(100) + (3 + bsqrt(2))^(100) = 7 + 5 sqrt(2)` `rArr (3 - a sqrt(2))^(100)+ (3 - b sqrt(2))^(100) = 7 - 5 sqrt(2)` We observe that LHS `gt` 0 for all pairs (a, b). So, the above equality is not ture for any pair (a, b) of ratinoal number. |
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| 64. |
If `(1 + x)^(n) = C_(0) + C_(1) x C_(2) x^(2) +…+ C_(n) x^(n)`, then the sum `C_(0) + (C_(0)+C_(1))+…+(C_(0) +C_(1) +…+C_(n -1))` is equal toA. `n 2^(n -1)`B. `( n-1)2^(n)`C. ` n 2 ^(n)`D. `(n+1)2^(n)` |
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Answer» Correct Answer - a We have, `C_(0) + (C_(0)+C_(1))+…+(C_(0) +C_(1) +C_(2)+…+C_(0)+C_(1) +C_(2)+...+ C_(n-1))` `nC_(0) +(n-1)C_(1) + (n-2) C_(2) +...+C_(n-1)` `= sum_(r=1)^(n)(n-r)""^(n)C_(r)` `= sum_(r=1)^(n)n. ""^(n)C_(r)- sum_(r=0)^(n) r. ""^(n)C_(r)` `=n sum_(r=1)^(n) ""^(n)C_(r)-n sum_(r=0)^(n) ""^(n-1)C_(r-1)= n2^(n) - n2^(n-1) = n2^(n-1)` . |
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| 65. |
So, statement-1 is also true. Stetement-2 is a correct expanation for statement-1. `S_(1)= sum_(j=1)^(10) j (j -1)""^(10)C_(j),S_(2)= sum_(j=1)^(10)j.""^(10)C_(j) and S_(2)= sum_(j=1)^(10)j.""^(10)C_(j) .` Statement-1 `S_(3) = 50xx2^(9)`. Statement-2 `S_(1) = 90xx2^(8) and S_(2) = 10 xx 2^(8)`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - d We have, `S_(2)= sum_(j=1)^(10) j ""^(10)C_(j)=sum_(j=1)^(10) jxx(10)/(j) xx(10)/(j) ""^(9)C_(j-1) = 10 sum_(j=1)^(10)""^(9)C_(j-1) = 10 xx2^(9)` `S_(1)= sum_(j=1)^(10) (j-1) ""^(10)C_(j)=sum_(j=1)^(10) j(j-1)xx(10)/(j) xx(9)/(j-1) xx""^(8)C_(j-2)` `rArr S_(2) = 10 xx9 sum_(j=2)^(10) ""^(8)C_(j-2) = 90 xx2^(8)` and, `s_(3) = sum_(j=1)^(10) j^(2) ""^(10)C_(j) = sum_(j=1)^(10){j (j -1)+ j} ""^(10)C_(j)` `rArr S_(3) = sum_(j = 1)^(10) j(j -1)""^(10)C_(j) + sum_(j=1)^(10) j ""^(10)C_(j) = 10 xx9 2^(8) + 10 xx2^(9)` `rArr S_(3) = 11 xx10 xx2^(8)` So, statement-2 is true and statement-1 is false. |
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| 66. |
The coefficient of `x^(20)` in ` (1 + 3x + 3x^(2) + x^(3))^(20)`,I sA. `""^(60)C_(40)`B. `""^(30)C_(20)`C. `""^(15)C_(2)`D. none of these |
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Answer» Correct Answer - a We have , `(1 + 3x + 3x^(2) + x^(3))^(20) = { (1+ x)^(3) }^(20) = (1 + x)^(60)` `therefore ` Coefficient of `x^(20) ` in` (1 + 3x + 3x^(2) + x^(3))^(20)` = Coefficient of `x^(20)` in `(1 + x)^(60) = ""^(60)C_(20) = ""^(60)C_(40)` |
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| 67. |
The sum of the series ` sum_(r=0) ^(n) ""^(2n)C_(r), ` isA. `n 2^(2n-1)`B. `2^(2n - 1)`C. `2^(n-1) + 1`D. none of these |
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Answer» Correct Answer - a We have, `rArr sum_(r=1)^(n)r. ""^(2n)C_(r) = sum_(r=1)^(n) r . (2n)/(r) ""^(2n-1)C_(r-1)` `rArr sum_(r=1)^(n)r. ""^(2n)C_(r) = 2n sum_(r=1)^(n) ""^(2n-1)C_(r-1)` `rArr sum_(r=1)^(n)r. ""^(2n)C_(r) = 2n {""^(2n-1)C_(0)+""^(2n-1)C_(1)+""^(2n-1)C_(2)+ ...+""^(2n-1)C_(r-1)}` `rArr sum_(r=1)^(n)r. ""^(2n)C_(r) = n {""^(2n-1)C_(0)+""^(2n-1)C_(1)+""^(2n-1)C_(2)+ ...+""^(2n-1)C_(r-1)+ ` `+ ""^(2n-1)C_(0)+""^(2n-1)C_(n+1)+...+""^(2n-1)C_(2n-2)+""^(2n-1)C_(2n-1)}` ""`[because ""^(n)C_(r) = ""^(n)C_(n-r)]` . `rArr sum_(r=1)^(n) r. ""^(2n)C_(r) = n (2(2n -1))` |
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| 68. |
Statement-1: `(C_(0))/(2.3)- (C_(1))/(3.4) +(C_(2))/(4.5)-.............+............+(-1)^(n) (C_(n))/((n+2)(n+3))= (1)/((n+1)(n+2))` Statement-2:`(C_(0))/(k)- (C_(1))/(k+1) +(C_(2))/(k+3)+............+(-1)^(n) (C_(n))/(k+n)=int_(0)^(1)x^(k-1) (1 - x)^(n) dx`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - a We have, `(1 - x)^(n) = C_(0) - C_(1) x + C_(2) x^(2) - ...+ (-1)^(n) C_(n)x^(4)` `rArr x^(k-1) (1 - x)^(n) = C_(0) x^(k-1) - C_(1) x^(k)+ C_(2) x^(k+1)-.... + (-1)""^(n)C_(n) x^(n+k-1) ` `rArr underset(0)overset(1)(int)x^(k-1) (1 - x)^(n) dx=(C_(0))/(k)-(C_(1))/(k+1) + (C_(2))/(k+2) - ....+ (-1)^(n) (C_(n))/(k+n)` So, statement-2 is true. Replacing k= 2 and k = 3 respecitvely, we get `underset(0)overset(1)(int)x^(k-1) (1 - x)^(n) dx=(C_(0))/(2)-(C_(1))/(3) + (C_(2))/(4) - ....+ (-1)^(n) (C_(n))/(n+2)` ...(i) and `underset(0)overset(1)(int)x (1 - x)^(n) dx=(C_(0))/(3)-(C_(1))/(4) + (C_(2))/(5) - ....+ (-1)^(n) (C_(n))/(n+2)` ...(ii) Subtracting (ii) from (i), we get `underset(0)overset(1)(int)x (1 - x)^(n+1) dx-(C_(0))/(2.3)-(C_(1))/(3.4) + (C_(2))/(4.5) - ....+ (-1)^(n) (C_(n))/((n+2)(n+3))` `rArr underset(0)overset(1)(int) (1 - x)x^(n+1) dx=(C_(0))/(2.3)-(C_(1))/(3.4) + (C_(2))/(4.5) - ...` `+ (-1)^(n) (C_(n))/((n+2)(n+3))[because underset(0)overset(a)(int) f(x) dx = underset(0)overset(a)(int)f(a-x)dx]` `underset(0)overset(1)(int) (x^(n+1)- x^(n+2)) dx=(C_(0))/(2.3)-(C_(1))/(3.4) + (C_(2))/(4.5)... + (-1)^(n) (C_(n))/((n+2)(n+3))` `rArr (1)/((n+1)(n+2))= (C_(0))/(2.3)-(C_(1))/(3.4) + (C_(2))/(4.5)-...+ (-1)^(n) (C_(n))/((n+2)(n+3))` Hence, statement-1 is also true and statement-2 is a correct expanation for statement-1 |
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| 69. |
The sum of the series ` sum_(r=0) ^(n) "r."^(2n)C_(r), ` isA. `2^(2n -1) `B. `n . 2^(2n-1)`C. `n 2 ^(n-1)`D. `n 2^(2n - 2)` |
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Answer» Correct Answer - b We have, `sum _(r=0) ^(n) r ""^(2n)C_(r) = sum_(r=0)^(n) r (2n)/(r) ""^(2n -1) C_(r -1)` `sum_(r=0) ^(n) r ""^(2n)C_(r) = n sum_(r=1)^(n) (""^(2n-1)C_(r-1)+""^(2n-1)C_(r-1))` `sum_(r=0)^(n) r ""^(2n)C_(r) = n sum_(r=0)^(n) (""^(2n-1)C_(r-1)+""^(2n - 1) C_(r-1))` `sum_(r=0)^(n) r ""^(2n)C_(r) = n sum_(r=0)^(2n-1) ""^(2n-1)C_(r)= n2^(2n -1)` |
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| 70. |
For any n `in` N, let `C_(r)` stand for `""^(n)C_(r)`, ` r = 0,1,2,3,…,n` and let `S= sum_(r=0)^(n) (1)/(C_(r))` Statement-1: `underset(0leilt i le n)(sumsum) ((i)/(C_(i))+(j)/(C_(j)))= (n^(n))/(2)S` Statement-2:` underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= nS`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - a We have, `underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= sum_(i=0)^(n-1) (n-i)/(C_(i) )+ sum_(j=1)^(n) (j)/(C_(j))` `rArr underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= n sum_(i=0)^(n-1) (1)/(C_(i) )- sum_(i=0)^(n) (i)/(C_(i)) + sum_(j=1)^(1)(j)/(C_(j))` `rArr underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= nsum_(i=0)^(n-1) (1)/(C_(i) )+(n)/(C_(n))` `rArr underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= nsum_(i=0)^(n-1) (1)/(C_(i) )=nS` So, statement-2 is true. Let `S_(1) underset(0leiltj len) (sumsum) ((i)/(C_(i))+(j)/(C_(j)))` . Then `S_(1) underset(0leiltj len) (sumsum) ((n-i)/(C_(n-i))+(n-j)/(C_(n-j)))` . `rArr S_(1) = n underset(0leiltj len) (sumsum) ((1)/(C_(n-i))+(1)/(C_(n-j)))- underset(0leiltj len) (sumsum)((i)/(C_(n-i))+(j)/(C_(n-j)))` `rArr S_(1) = n underset(0leiltj len) (sumsum) ((1)/(C_(i))+(1)/(C_(j)))- underset(0leiltj len) (sumsum)((i)/(C_(i))+(j)/(C_(j)))` `rArr S_(1) = n(nS) - S_(1)` [ Using truth of statement-2] `rArr 2S_(1) = n^(2) S rArr S_(1) = (n^(2))/(2)S` So, statement-1 is true and stetement-2 is a correct expanation for statement-1. |
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| 71. |
The sum of the series ` sum_(r=0) ^(n) ""^(2n)C_(r), ` isA. `2^(2n)`B. `2^(n)`C. `2^(2n) + ""^(2n)C_(n)`D. `(1)/(2) (2^(2n) + ""^(2n)C_(n))` |
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Answer» Correct Answer - d We have, `sum_(r=0)^(n) ""^(2n)C_(r) = (1)/(2)sum_(r=0)^(n) (2^(2n) + ""^(2n)C_(n))` ltbbrgt `rArr sum_(r=0)^(n) ""^(2n)C_(r) = (1)/(2)sum_(r=0)^(n) ""^(2n)C_(2n) + ""^(2n)C_(n)r = (1)/(2)[ {sum_(r=0)^(2n) ""^(2n)C_(r)}+""^(2n)C_(n)r]` `rArr sum_(r=0)^(n) ""^(2n)C_(r) = (1)/(2) (2^(2n) + ""^(2n) C_(n))` . |
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| 72. |
Statement -1: ` sum_(r=0)^(n) r(""^(n)C_(r))^(2) = n (""^(2n -1)C_(n-1))` Statement-2: ` sum_(r=0)^(n) (""^(n)C_(r))^(2)= ""^(2n)C_(n)`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - b For truth of statement-2, see illustration 17 on page 14,16. Now, `sum_(r=0)^(n) rC_(r)""^(2)= sum_(r=0)^(n) r .""^(n)C_(r).""^(n)C_(r).=` `rArr sum_(r=0)^(n) rC_(r)""^(2)=n sum_(r=0)^(n) ""^(n-1)C_(r-1).""^(n)C_(r).` `rArr sum_(r=0)^(n) rC_(r)""^(2)=n .sum_(r=0)^(n) ""^(n-1)C_(n-r).""^(n)C_(r).` `rArr sum_(r=0)^(n) rC_(r)""^(2)=n` Coefficient of `x^(n)` in `{ (1 + x)^(n -1) (1 + x)^(n)}` `rArr sum_(r=0)^(n) rC_(r)""^(2)=n` Coefficient of `x^(n)` in ` (1 + x)^(2n+1)` `rArr sum_(r=0)^(n) rC_(r)""^(2)=n^(2)""^(n-1)C_(n) = n^(2)""^(n-1)C_(n-1)`. So, statement-1 is also true. But , statement-2 is not a correct expanation for statement -1. |
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| 73. |
The value of `1^2.C_1 ``+ 3^2.C_3 +`` 5^2.C_5 + ...` isA. `n(n -1)2^(n-2) + n.2^(n-1)`B. `n(n -1)2^(n-2)`C. `n(n+1). 2 ^(n-3)`D. none of these |
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Answer» Correct Answer - c We have, `sum_(r=1) ^(n) r^(2) ""^(n)C_(r) = n (n-1)2^(n-2) + n.2 ^(n-1)` and , `sum_(r=1)^(n) (-1)^(r-1) r^(2) ""^(n)C_(r) = 0` Adding these two, we get `2{1^(2) C_(1) + 3^(2) C_(3) + 5^(2) C_(5) + …} = n (n-1) 2^(n-2) + n. 2^(n-1)` `rArr 1^(2) C_(1) + 3^(2) . C_(3) + 5^(2). C_(5) +...= n(n -1) 2^(n-3) + n. 2^(n -2)` . |
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| 74. |
The value of `(sumsum)_(0leilejlen) (""^(n)C_(i) + ""^(n)C_(j))` is equal toA. `n2^(n)`B. `(n+1)2^(n)`C. `(n +2) 2^(n)`D. none of these |
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Answer» Correct Answer - c We have, `underset(0leilejlen)(sumsum)(""^(n)C_(i) + ""^(n)C_(j))` ltbrge `underset(0leilejlen)(sumsum)""^(n)C_(i) +underset(0leilejlen)(sumsum) ""^(n)C_(j)` = `sum_(i=0)^(n) (n - i +1)""^(n)C_(i) + sum_(i=0)^(n) (j +1) ""^(n)C_(j)` = `sum_(i=0)^(n) (n -r+1)""^(n)C_(r)+ sum_(i=0)^(n) (r +1)""^(n)C_(r)` `sum_(i=0)^(n) (n +2)""^(n)C_(r)= (n+2) sum_(i=0)^(n) ""^(n)C_(j)= (n+2)2^(n)`. |
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| 75. |
The remainder when `9^103` is divided by 25 is equal toA. 5B. 6C. 4D. none of these |
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Answer» Correct Answer - c We have, `9^(103)` =` 9xx(81)^(51)` = ` 9(80 + 1)^(51)` `= 9{""^(51) C_(0) (80)^(51)+ ""^(51)C_(1) (80)^(50) + ""^(51)C_(2)(80)^(49) +...+""^(51)C_(50)(80) +1}` `9 [ 25 k + (51 xx80 + 1)] ` `= 25 k lambda+ 36729` `25 lambda + 25 xx1469 + 4` ` 25(lambda + 14 69) + 4` Hence, the remainder is 4, |
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| 76. |
`2^60` when divided by 7 leaves the remainderA. 1B. 6C. 5D. 2 |
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Answer» Correct Answer - A We have, `2^(60) = (1 + 7)^(20)` `rArr 2^(60) = ""^(20)C_(0) + ""^(20) C_(1)xx 7 + ""^(20)C_(2)xx7^(2) + …+ ""^(20)C_(20) xx7^(20)` `rArr 2^(60) =7( ""^(20)C_(1) + ""^(20) C_(2)xx 7 +...+ ""^(20)C_(2)xx7^(19))+1`. `2^(60)` when divided by 7 leaves the remainder 1. . |
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| 77. |
If `sum_(i=1)^(n-1) ((""^(n)C_(i-1))/(""^(n)C_(i)+""^(n)C_(i-1)))^(3) = (36)/(13)`, , then n is equal toA. 10B. 11C. 13D. 12 |
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Answer» Correct Answer - d We have , `sum_(i-1)^(n-1) ((""^(n)C_(i-1))/(""^(n)C_(i)+""^(n)C_(i-1)))^(3) = (36)/(13)`, , `rArr sum_(i-1)^(n-1) ((""^(n)C_(i-1))/(""^(n+1)C_(i)))^(3) = `(36)/(13) ` [ `because ""^(n) C_(r-1) + ""^(n)C_(r) = ""^(n+1)C_(r)]` `rArr sum_(i-1)^(n-1) ((""^(n)C_(i-1))/((n+1)/(i)""^(n)C_(i-1)))^(3) = (36)/(13) ` `rArr sum_(i-1)^(n-1) ((i)/(n+1))^(3) = (36)/(13)` `rArr (1)/(n+3)^(3) {(n(n+1))/(2)}^(2) = (36)/(13)` `(n^(2))/(4 (n +1)^(3))= (36)/(13)` `rArr (n^(2))/ (n +1)=(144)/(13) rArr (n^(2))/(n+1) = (12^(2))/((12 +1))rArr n=12`. |
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| 78. |
The vaule of ` sum_(r=0)^(n-1) (""^(n)C_(r))/(""^(n)C_(r) + ""^(n)C_(r +1))` is equal toA. `(n-1)/(n+1)`B. `(n+1)/(2)`C. `(n(n+1))/(2)`D. `(n)/(2)` |
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Answer» Correct Answer - d We have ` sum_(r=0)^(n-1) (""^(n)C_(r))/(""^(n)C_(r) + ""^(n)C_(r+1)) ` ` sum_(r=0)^(n-1) (""^(n)C_(r))/( ""^(n+1)C_(r+1)) ` ` sum_(r=0)^(n-1) (""^(n)C_(r))/( (n+1)/(r +1)""^(C_(r))) ` ` sum_(r=0)^(n-1) (r+1)/(r+1)=(1)/(n+1)sum_(r=0)^(n-1) ( r+1) = (1)/(n+1) xx(n (n+1))/(n +1)= (n)/(2).` . |
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| 79. |
The value of `sum_(r=0)^(15)r^(2)((""^(15)C_(r))/(""^(15)C_(r-1)))` is equal toA. 1085B. 560C. 680D. 1240 |
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Answer» Correct Answer - c We have, ` ((""^(15)C_(r))/(""^(15)C_(r-1)))= (n-r +1)/(r)` `therefore sum_(r=1)^(15) r^(2) ((""^(15)C_(r))/(""^(15)C_(r-1)))sum_(r=1)^(15) r^(2)((15 - r +1)/(r)) = sum_(r=1)^(15) r (16 -r)` `= 16 sum_(r=1)^(15) r- sum_(r=1)^(15) r^(2)` `= (16xx15xx15)/(2) - (15xx16xx31)/(6) = 680`. |
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| 80. |
The valur of `sum_(r=0)^(n) sum_(p=0)^(r) ""^(n)C_(r) . ""^(r)C_(p)` is equal toA. `3^(n) - 2^(n)`B. `3^(n) - 2^(n) - 2`C. `3^(n) - 2^(n) + 2`D. none of these |
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Answer» Correct Answer - d We have , `sum_(r=0)^(n) sum_(r=0)^(n) ""^(n)C_(r) . ""^(r)C_(p)` `sum_(r=0)^(n) ""^(r)C_(p) (sum_(r=0)^(n) ""^(n)C_(p)) = sum_(r=0)^(n) . ""^(n)C_(r) 2^(r) = (1 +2)^(n) = 3^(n)` . |
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| 81. |
`7^(103)` when divided by 25 leaves the remainder .A. 20B. 16C. 18D. 15 |
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Answer» Correct Answer - c We have, ` 7^(103) = 7(49)^(51)` `rArr 7^(103) = 7(50 -1)^(51)` ` rArr 7^(103) = 7{""^(51)C_(90)(50)^(51) - ""^(51)C_(r)(50)^(50) + ""^(51)C_(2)(50)^(49)...-1}` ` rArr 7^(103) = 7{(50)^(51) - ""^(51)C_(1)(50)^(50) + ""^(51)C_(2)(50)^(49)...}-7` `rArr 7^(103) = 7{(50)^(51) - ""^(51)C_(1)(50)^(50) + ""^(51)C_(2)(50)^(49)...}-7 - 18 + 18` `rArr 7^(103) = 7{(50)^(51) - ""^(51)C_(1)(50)^(50) + ""^(51)C_(2)(50)^(49)...}-25 + 18` `rArr 7^(103)` = Multiple of 25 + 18 Hence, remainder = 18. |
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| 82. |
If `(1 + x + x^(2) + x^(3))^(n)= a_(0) + a_(1)x + a_(2)x^(2) + a_(3) x^(3) +...+ a_(3n) x^(3n)`, then the value of `a_(0) + a_(4) +a_(8) + a_(12)+….. ` isA. -1B. 0C. `4^(n-1)`D. n |
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Answer» Correct Answer - c We have, ` a_(0) + a_(1)x + a_(2)x^(2) + a_(3) x^(3) +...+ a_(3n) x^(3n)=(1 + x + x^(2) + x^(3))^(n)` `rArr a_(0) + a_(2)x^(2) + a_(4)x^(4) +...)+( a_(1) x + a_(3) x^(3)+...) =(1 + x + x^(2) + x^(3))^(n)` Puttinhg x = 1, - 1, i and -i respectively, we get `( a_(0) + a_(2) + a_(4) +...)+( a_(1) + a_(3) +a_(5)+...) =4^(n)` ....(i) `( a_(0) + a_(2) + a_(4) +...) - ( a_(1) + a_(3) +a_(5)+...) = 0 ` ...(ii) `( a_(0) - a_(2) + a_(4) - a_(6) +...)+i ( a_(1) - a_(3) +a_(5)+...) = 0 ` ...(iii) `( a_(0) - a_(2) + a_(4) - a_(6) +...)- i ( a_(1) - a_(3) +a_(5)+...) = 0 ` ....(iv) Adding (i) and (ii), we get ` 2 (a_(0) + a_(2) + a_(4) + ....) = 4 ^(n)` ....(v) Adding (iii) and (iv), we get `2 (a _(0) - a_(2) + a_(4) - a_(6)+ ....) = 0` ...(vi) Adding (v) and (vi), we get ` 4 ( a_(0) + a_(4) + a_(8) + ....) = 4^(n) rArr a_(0) + a_(4) + a_(8) + ... = 4^(n-1)` |
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| 83. |
Let m be the smallest positive integer such that the coefficient of `x^2` in the expansion of `(1+x)^2 + (1 +x)^3 + (1 + x)^4 +........+ (1+x)^49 + (1 + mx)^50` is `(3n + 1) .^51C_3` for some positive integer n. Then the value of n isA. 16B. 5C. 21D. 11 |
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Answer» Correct Answer - b We find that `(1 + x)^(2) + (1 + x)^(3) +… (1 + x)^(49) + (1 + mx)^(50)` ` = (1 + x)^(2) { ((1 + x)^(48) -1)/((1 + x) -1)} +(1 + mx)^(50)` ` (1)/(x) {( 1 + x)^(50) - (1 + x)^(2)} + (1 +mx)^(50)` Coeffi. Of `x^(2)` in `(1 + x)^(2) + (1 + x)^(3) + ...(1 + x)^(49) + (1 +mx)^(50)` = Coeffi.of `x^(2)` in `[(1)/(x) {(1 +x)^(50) - (1 + x)^(2)} +(1 +mx)^(50)]` ` Coeffi. of `x^(3)` in `{ (1 + x)^(50) - (1 + x)(2)} + ` Coefficient of `x^(2)` in `(1 +mx)^(50)` `""^(50)C_(3) + ""^(50)C_(2) m^(2)` It is given that this coefficient is `(3n +1) ""^(51)C_(3)` . `therefore ""^(50)C_(3) + ""^(50)C_(2) m^(2) = (3n +1)""^(51)C_(3)` `rArr (50xx49xx48)/(3xx2xx1)+(50xx49)/(2xx1)m^(2) = (3n-1)(51xx50xx49)/(3xx2xx1)` `rArr 16 + m^(2) = 17 (3n +1)` `rArr m^(2) = 51n+1` The least value of n for which `51 N+1` is a perfect square is 5 and for this value of n the value of m is 16. Hence,m=16 and n=5. |
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| 84. |
The value of `sum_(r=1)^(10) r. (""^(n)C_(r))/(""^(n)C_(r-1)` is equal toA. ` 5 (2n -9)`B. `10`nC. `9 (n-4)`D. none of these |
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Answer» Correct Answer - a We have , `sum_(r=1)^(10)r. (""^(n)C_(r))/(""^(n)C_(r-1)) = sum_(r=1)^(10) (n-r+1) ` `rArr sum_(r=1)^(10)r. (""^(n)C_(r))/(""^(n)C_(r-1)) = sum_(r=1)^(10) {(n+1) - r}` ` rArr sum_(r=1)^(10)r. (""^(n)C_(r))/(""^(n)C_(r-1)) = 10 (n+1) - sum_(r=1)^(10) r` `rArr sum_(r=1)^(10)r. (""^(n)C_(r))/(""^(n)C_(r-1)) = 10 (n+1) -55= 10 n -45 = 5 (2n - 9).` |
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| 85. |
Let `(1 + x)^(36) = a_(0) +a_(1) x + a_(2) x^(2) +...+ a_(36)x^(36)`. Then Statememt-1: `a_(0) +a_(3) +a_(6) +…+a_(36)= (2)/(3) (2^(36) +1)` Statement-2: `a_(0) + a_(2) +a_(4) +…+ a_(36) = 2^(35)`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - b We have, `a_(0) + a_(1) + a_(2) x^(2) + a_(3) + … + a_(36) x^(36) = (1 + x)^(36)` …(i) Putting x = 1 `omega, omega^(2)` successively, we get `(a_(0) + a_(3) + a_(6) +...+ a_(36) )+ (a_(1) + a_(4) + a_(7) +...+a_(34))+(a_(2) + a_(5) +...+a_(35)) = 2^(36)` ...(ii) `(a_(0) + a_(3) + a_(6) +...+ a_(36) )+omega (a_(1) + a_(4) + a_(7) +...+a_(34))+omega^(2)(a_(2) + a_(5) +...+a_(35)) = 1` ...(iii) `(a_(0) + a_(3) + a_(6) +...+ a_(36) )+omega^(2) (a_(1) + a_(4) + a_(7) +...+a_(34))+omega(a_(2) + a_(5) +...+a_(35)) = 1` `(a_(0) + a_(3) + a_(6) +...+ a_(36) )+omega^(2) (a_(1) + a_(4) + a_(7) +...+a_(34))+omega(a_(2) + a_(5) +...+a_(35)) = 1` ....(iv) Adding these three , we get `3(a_(0) + a_(3) + a_(6) +....+ a_(36)) = 2 (2^(35) +1)` `rArr a_(0) + a_(3) + a_(6) +....+a_(36) = (2)/(3) (2^(35) +1)` So, statement-1 is true. Putting x = -1 in (i) and adding it to (ii) , we get `a_(0) + a_(2) + a_(4) +...+ a_(36) = 2^(35)` So, statement-2 is true. But , statement-2 is not a correct explanation for satatement -1 . |
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| 86. |
FOr ` xin r, x != -1` , If `(1+x)^(2016)+x(1+x)^(215)+x^2(1+x)^(2014)++x^(2016)=sum_(i=0)^2016 a_i x^i`, then `a_17` is equal to -A. `(2016!)/(16!)`B. `(2017!)/(2000!)`C. `(2017!)/(17!2000!)`D. `(1016!)/(17!1999!)` |
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Answer» Correct Answer - c We have, `(1 + x)^(2016) + x(1 + x)^(2015) + x^(2) (1 + x)^(2014) + ..+ x^(2016) = sum_(i=0)^(2016) a_(i) x^(i)` `rArr (1 +x)^(2016) {(((x)/(1|x))^(2017) -1)/((x)/(1+x) -1)}= sum_(i=1)^(2016) a_(1) x^(j) ` `rArr {(1 + x)^(2017) - x ^(2017)} = sum_(i=1)^(2016) a_(i) x^(j)` `rArr a_(17) = ` Coefficient of `x^(17)` in ` {( 1 + x)^(2017) - x^(2017)}` `rArr a_(17)` = Coefficient of `x^(17)` in `(1 + x)^(2017)` `rArr a_(17) = ""^(2017)C_(17) = (2017!)/(17!2000!)` |
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| 87. |
P is a set containing n elements . A subset A of P is chosen and the set P is reconstructed by replacing the element of A. B such that A and B have no common elements, isA. `2^(n)`B. `3^(n)`C. `4^(n)`D. none of these |
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Answer» Correct Answer - b Suppose A contains ` r (0 le r len)` elements. Then , B is constructed by selecting some elements from the remaining `(n-r)` elements. Clearly, A can be chosen in `""^(n)C_(r)` ways and B in `(""^(n-r)C_(0)+""^(n-r)C_(1)+...+""^(n-r)C_(n-r) )= 2^(n-r)` ways.So, total number of ways of choosing A and B is `""^(n)C_(r) xx2^(n-r)`. But , r can very from 0 to n. `therefore ` Required number of ways = `sum_(r=0)^(n) ""^(n)C_(r) xx2^(n-r) = (1 + 2)^(n) = 3^(n)` . |
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| 88. |
`(1 + x + x^(2))^(n) = a_(0) +a_(1)x + a_(2) x^(2) +...+ a_(2n )x^(2n)` , then `a_(0) a_(2n) - a_(1) a_(2r+1) + a_(2) a_(2r+2) - a_(3) a_(2r+3) +...+a_(2n-2r)a_(2n)=` . |
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Answer» Correct Answer - d On equating the coefficients of `x^(2n +1)` on both sides of (v) in illustration 25, we get `a_(0) a_(2n) - a_(1) a_(2r+1) + a_(2) a_(2r+2) - a_(3) a_(2r+3) +...+a_(2n-2r)a_(n+r)`. |
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| 89. |
There are p letters a, q letters b, r letters C.The number of ways of selecting k letters out of these if `p lt k lt q lt r` isA. `3^(n-1)`B. `n 3^(n)`C. `n3 ^(n-1)`D. `2 n ^(n-1)` |
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Answer» Correct Answer - c If P `cup` Q conains exaclty one element , both P and Q must be non-enpty. Thus, if P has r elements, Q must have exactly one of these r elements and any number of elements from among the reamaining (n-r ) elements in A, so that the number of ways of choosing Q is `""^(r)C_(1) xx2^(n-r)` . But, P can be chosen in `""^(n)C_(r)` ways. Therefore, P and Q can be chosen in `""^(n)C_(1) xx""^(r)C_(1)xx2^(n-r)` , when P contains r elements. As r can very from 1 to n . Therefore, P and Q in general can be chosen in `sum_(r=1)^(n) ""^(n)C_(r) xx""^(n)C_(r)xx2^(n-r)` `=sum_(r=1)^(n) ""^(n)C_(r)r2^(n-r)` `=sum_(r=1)^(n) (n)/(r) ""^(n-r)C_(r-1).r.2^(n-r)` "" `[because ""^(n)C_(r)= (n)/(r) ""^(n-1)C_(r - 1)]` `nsum_(r=0)^(n) ""^(n-1)C_(r-1)2^(n-r)` `nsum_(r=0)^(n) ""^(n-1)C_(r-1)2^((n-r)-(r-0)) = n(1 + 2)^(n+1) = n3^(n-1)` |
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| 90. |
A is a set containing n elements . A subset of A is chosen at random. The set A is reconstructed by replacing the elements of P.A subet Q is again chosen at random. The number of ways of selecting P and Q, is .A. `2^(n)`B. `4^(n)`C. `2n`D. `3^(n)` |
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Answer» Correct Answer - b The set A has n elemeats. So, it has `2^(n)` sunbsets. Therefore, set P can be chosen in`""^(2^(n))C_(1)` ways. Similarly, set Q can also be chosen in `""^(2^(n))C_(1)` ways . `because ` Sets P and Q can be chosen in `""^(2^(n))C_(1)xx""^(2^(n))C_(1)=2^(n) xx2^(n) = 4^(n)` ways. |
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| 91. |
`sum_(r=0)^n (-1)^r .^nC_r (1+rln10)/(1+ln10^n)^r`A. 1B. -1C. nD. none of these |
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Answer» Correct Answer - d Let `log_(e) 10 = x` . Then `sum_(r=0)^(n) (-1)^(r) ""^(n)C_(r) (1 + r log_(e)10)/((1 + log_(e) 10^(n))^(r) ) ` `sum_(r=0)^(n) (-1)^(r) ""^(n)C_(r)(1 + rx)/((1 + nx)^(r))` `sum_(r=0)^(n) (-1)^(r) ""^(n)C_(r)((1 )/(1 + nx))^(r)+sum_(r=0)^(n) (-1)^(r) (n)/(r) ""^(n-1)C_(r-1) (rx)/((1 + nx)^(r))` `sum_(r=0)^(n) (-1)^(r) ""^(n)C_(r)((1 )/(1 + nx))^(r)- (nx)/(1 + nx)sum_(r=1)^(n) (-1)^(r-1) ""^(n-1)C_(r-1) (1)/((1 + nx)^(r))` ` = (1 - (1) /(1+ nx))^(n) - (nx)/(1 + nx) (1 - (1)/(1 + nx))^(n-1)` `= ((nx)/(1+nx))^(n) = ((nx)/(1+nx))^(n) = 0` . |
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| 92. |
If `f (n) = sum_(s=1)^n sum_(r=s)^n "^nC_r`` "^rC_s` , then `f(3) =`A. 27B. 19C. 1D. 5 |
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Answer» Correct Answer - b We have, `f(n) = sum_(s=1)^(n) sum_(r=s)^(n) ""^(n)C_(r) ""^(r)C_(s)` `rArr f(n) = sum_(r =1)^(n) ""^(n)C_(1) + sum_(r=2)^(n) ""^(n)C_(r) ""^(r)C_(2) +….+ sum_( r= n -1)^(n) ""^(n)C_(r) ""^(r)C_(n-1) + ""^(n)C_(n) ""^(n)C_(n) ` `rArr f(n) = ""^(n)C_(1)""^(1)C_(1)+ ""^(n)C_(2)(""^(2)C_(1) + ""^(2)C_(2)) + ""^(n)C_(3) (""^(3)C_(1) + ""^(n)C_(2))+""^(n)C_(3)(""^(3)C_(1)+""^(3)C_(2) + ""^(3)C_(n))` ` + ....+ ""^(n) C_(n) (""^(n)C_(1) + ""^(n)C_(2) +.... + ""^(n)C_(n))` `rArr f(n) = ""^(n)C_(1)xx(2^(1) -1) + ""^(n)C_(2) xx(2^(2) -1) + ""^(n)C_(3) xx(2^(3) -1) +... + ""^(n)C_(n) xx(2^(n) -1)` `rArr f(n) = sum_(r=1)^(n) = ""^(n)C_(r) 2^(r) - sum_(r=1)^(n) ""^(n)C_(r)` `rArr f(n) = sum_(r=0)^(n) = ""^(n)C_(r) 2^(r) - sum_(r=1)^(n) ""^(n)C_(r)` `rArr f(n) = 3^(n) -2^(n)` `rArr f(3) = 3^(3) - 2^(3) = 19` |
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| 93. |
If n is an odd natural number, then `sum_(r=0)^n (-1)^r/(nC_r)` is equal to |
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Answer» Correct Answer - a We have, `sum_(r=0)^(n) ((-1)^(r))/(""^(n)C_(r) )= sum_(r=0)^((n+1)/(2)) {(( -1)^(r))/(""^(n)C_(r) )+ (-1)^(n-r)/(""^(n)C_(n-r))}` `rArr sum_(r=0)^(n) ((-1)^(r))/(""^(n)C_(r) )= sum_(r=0)^((n+1)/(2))(-1)^(r) {(( -1)^(r))/(""^(n)C_(r) )+ (-1)^(n-r)/(""^(n)C_(n-r))}` `rArr sum_(r=0)^(n) ((-1)^(r))/(""^(n)C_(r) )= sum_(r=0)^((n+1)/(2))(-1)^(2) {( 1)/(""^(n)C_(r) )+ (1)/(""^(n)C_(r))}= 0 [{:(therefore " is odd and "),(""^(n)C_(r) = ""^(n)C_(n-r)):}]` . s |
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| 94. |
If `S_n=sum_(r=0)^n 1/(nC_r) and sum_(r=0)^n r/(nC_r),` then `t_n/S_n=`A. `(2 n - 1)/(2) `B. `(n)/(2) -1`C. n-1D. `(n)/(2)` |
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Answer» Correct Answer - d We have, `s_(n) = sum_(r=0)^(n) (r)/(""^(n)C_(r)) = (n)/(2) s_(n)` [ See Example 37] `rArr (t_(n))/(s_(n)) = (n)/(2)` . |
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| 95. |
If `s_n=sum_(r < s) (1/(nC_r)+1/(nC_s)) and t_n=sum_(r < s)(r/(nC_r)+s/(nC_s)),` then `t_n/s_n=`A. n-1B. n+1C. `(n)/(2)`D. none of these |
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Answer» Correct Answer - c We have, `t_(n) =sum_(rlts) ( (r)/(""^(n)C_(r))+(s)/(""^(n)C_(s)))` `rArr t_(n) =sum_(rgts)( (r)/(""^(n)C_(r))+(s)/(""^(n)C_(s)))` `rArr t_(n) =sum_(n-rltn-s)( (n-r)/(""^(n)C_(n-r))+(n-s)/(""^(n)C_(n-s)))` `rArr t_(n) =sum_(rlts)( (1)/(""^(n)C_(r))+(1)/(""^(n)C_(s)))- sum_(rlt s)( (r)/(""^(n)C_(r))+(s)/(""^(n)C_(s)))` `rArr t_(n) = n s_(n) - t_(n) rArr (t_(n))/(s_(n)) = (n)/(2)` |
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| 96. |
Let `f (n) =sum_(r=1)^n r^2 C_r^2`.Then, `f (5) =`A. `""^(8)C_(4)`B. `25xx""^(8)C_(4)`C. `25 xx""^(8)C_(5)`D. none of these |
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Answer» Correct Answer - b `f(n) = sum_(r=1)^(n) r^(2) C_(r)""^(2)` `rArr f(0)= sum_(r=1)^(n) (r ""^(n)C_(r))^(2) = n^(2) sum_(r=1)^(n) (""^(n-1)C_(r-1))^(2)` `rArr f(n) = n^(2)""^(2n -2)C_(n-1)` `rArr (5) = 25xx""^(8)C_(4)` |
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| 97. |
The coefficient of `x^(7)` in the expansion of `(1-x-x^(2) + x^(3))^(6)` is :A. 144B. -132C. -144D. 132 |
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Answer» Correct Answer - c We have `(1 - x - x^(2) + x^(3))^(6) = {(1 - x)( 1 - x^(2))}^(7) = (1 - x )^(7) = (1 - x)^(7) (1 - x^(2))^(7)` = `(""^(6)C_(0) - ""^(6)C_(1) x + ""^(6)C_(2)x^(2) - ""^(6)C_(3)x^(3) +... + ""^(6)C_(6)x^(6))` `(""^(6)C_(0) - ""^(6)C_(1) x^(2) + ""^(6)C_(2)x^(4) - ""^(6)C_(3)x^(6) +... )` `therefore` Coefficient of `x^(7)` in `(1 - x - x^(2) + x^(3))^(6)` ` = - ""^(6)C_(1)xx (-""^(6)C_(3)) - ""^(6)C_(3)xx ""^(6)C_(3)xx ""^(6)C_(2) - ""^(6)C_(5) xx - (""^(6)C_(1))` ` 120 - 300 + 36 = -144` . |
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| 98. |
If `sum_(r=0)^(2n)(-1)^r(2n(C_r)^2=alpha_1`, then `sum_(r=0)^(2n)(-1)^r(r-2n)(2n(C_r)^2=`A. `nalpha`B. `-n alpha `C. `0D. none of these |
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Answer» Correct Answer - b Let `beta = sum_(r=0)^(2n) (-1)^(r) (r -2n)(""^(2n)C_(r))^(2)` …(i) `rArr beta = - sum_(r=0)^(2n) (-1)^(r) (2n -r) (""^(2n)C_(2n-r))^(2)` Writing the terms in the neverse order, we get `beta = sum_(r=0)^(2n) (-1)^(r) r(""^(2n)C_(r))^(2) ` ...(ii) Adding (i) and (ii), we get `2beta = - sum_(r=0)^(2n) (-1)^(r) (""^(2n)C_(r))^(2)` `rArr 2 beta = - 2n sum_(r=0)^(2n) (-1)^(r) (""^(2n)C_(r))^(2)` `rArr 2 beta = - 2nalpha rArr beta = - n alpha ` . |
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| 99. |
For `r= 0, 1,.....,10`, let `A_r,B_r, and C_r`denote, respectively, the coefficient of `x^r` in the expansions of `(1 +x)^10,(+x)^20 and (1+ x)^30`.Then `sum_(r=1)^10 A_r(B_10B_r-C_10A_r)` is equal toA. `B_(10 - C_(10)`B. `A_(10)(B_(10)^(2) - C_(10)A_(10))`C. 0D. `C_(10)- B_(10)` |
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Answer» Correct Answer - d We have, `A_(r) = ""^(10)C_(r), B_(r) = ""^(20)C_(r) and C_(r) = ""^(30)C_(r)` `because sum _(r=1)^(10) A_(r) (B_(10)B_(r)- C_(10)A_(r))` `= B_(10) sum_(r=1)^(10) A_(r) B_(r) - C_(10) sum_(r=1)^(10) (A_(r))^(2)` `= B_(10)(sum_(r=1)^(10) ""^(10)C_(r) ""^(20)C_(r)) - C_(10) sum_(r=1)^(10) (""^(10)C_(r))^(2)` `= B_(10)(sum_(r=1)^(10) ""^(10)C_(r) ""^(20)C_(20-r)) - C_(10) {sum_(r=1)^(10) (""^(10)C_(r))^(2) -1}` `= B_(10){sum_(r=1)^(10) ""^(10)C_(r) ""^(20)C_(20-r)-1} - C_(10) {sum_(r=1)^(10) (""^(10)C_(r))^(2) -1}` ` b_(10)XX` {Coefficient of `x^(20)` in `(1 + x )^(10) (x + 1) ^(20) -1} - C_(10) {""^(20)C_(10 -1)}` `B_(10)xx {""^(30) C_(20) -1} - C_(10) {""^(20)C_(10) -1}` `B_(10){ ""^(30)C_(10) -1} - C_(10) {""^(20)C_(10) -1} ` `B_(10) (C_(10) -1} - C_(10) { ""^(20) C_(10) -1}`. |
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| 100. |
Find the sum of the coefficients of all the integral powers of `x`in the expansion of `(1+2sqrt(x))^(40)dot`A. `3^(40) +1`B. `3^(40) -1`C. `(1)/(2)(3^(40) -1)`D. `(1)/(2)(3^(40) +1)` |
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Answer» Correct Answer - d We have, `( 1 + 2sqrt(x))^(40)= ""^(40)C_(0) + ""^(40)C_(1)(2sqrt(x)) + ""^(40)C_(2)(2sqrt(x))^(2) + ...+ ""^(40)C_(40)(2sqrt(x))^(40)` `rArr ( 1 + 2sqrt(x))^(40)= (""^(40)C_(0) + ""^(40)C_(1)xx2x+ ""^(40)C_(4)xx2^(2) x^(2) + ...+ ""^(40)C_(20)xx 2^(10)x^(10))` +`(""^(40)C_(1) xx2sqrt(x)+ ""^(40)C_(3)xx2 + ""^(40)C_(3)xx2^(2) + ...+ ""^(40)C_(19)xx 2^(19)x^(10//2))` ...(i) Putting `sqrt(x) = 1 ` and - 1 respectively, we get `3^(40) = {""^(40)C_(0)+ ""^(40)C_(2)xx2 + ""^(40)C_(4)xx2^(2) + ...+ ""^(40)C_(20)xx 2^(10)}` `+{""^(40)C_(1) xx 2 + ""^(40)C_(3)xx2^(3) + ...+ ""^(40)C_(20)xx2^(19)}` and , `1= {""^(40)C_(0) + ""^(40)C_(2)xx2 +""^(40)C_(4)xx2^(2)+...+""^(40)C_(20)xx2^(10)}` `- {""^(40)C_(1)xx2 + ""^(40)C_(3)xx2^(3) +...+""^(40)C_(19)xx2^(10)}` `therefore 3^(40) + 1 = 2 {""^(40)C_(0)+ ""^(40)C_(2)xx2+ ""^(40)C_(4) xx2^(2)+...+""^(40)C_(20)xx2^(10)}` `rArr ""^(40)C_(0)+ ""^(40)C_(2)xx2+ ""^(40)C_(4) xx2^(2)+...+""^(40)C_(20)xx2^(10)= (3^(40) + 1)/(2)` . |
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