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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. |
If `x`is so small that `x^3`and higher powers of `x`may be neglectd, then `((1+x)^(3//2)-(1+1/2x)^3)/((1-x)^(1//2))`may be approximated as`3x+3/8x^2`b. `1-3/8x^2`c. `x/2-3/xx^2`d. `-3/8x^2`A. `3x+3/(x^(2))`B. `1-3/8 x^(2)`C. `x/2-3/x x^(2)`D. `-3/8 x^(2)` |
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Answer» Correct Answer - D `((1-x)^(3//2)(1+(1)/(2)x)^(3))/((1-x)^(1//2))=((1+3/2x+3/8x^(2))-(1+3/2x+3(x^(2))/(4)))/((1-x)^(1//2))` `= (-3)/(8)x^(2)(1-x)^(-1//2)` `= - 3/8 x^(2)(1+x/2) = - 3/8 x^(2)` |
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| 52. |
If `x`is positive, the first negative term in the expansion of `(1+x)^(27//5)i s(|x| |
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Answer» Correct Answer - B `T_(r+1)` in `(1+x)^(n)` is `(n(n-1)(n-2)"……"(n-r+1))/(r!) x^(r)` For first negative term, `n -r + 1 lt 0` or `27/5 - r +1 lt 0` or `r gt 32/5` Thus, first negative term occurs when `r = 7`, |
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| 53. |
Find the value of `(sumsum)_(0leiltjlen) (1+j)(""^(n)C_(i)+""^(n)C_(j))`. |
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Answer» Here sum does nto change if we replace `i` by `n-1` and j by `n-j`. By doing so, in fact we are writing the series in the reverse order. Therefore, `S = underset(0leiltjlen)(sumsum)(i+j)(.^(n)C_(i)+.^(n)C_(j))" "(1)` `=underset(0leiltjlen)(sumsum)(n-i+n-j)(.^(n)C_(n-i)+.^(n)C_(n-j))` `=underset(0leiltjlen)(sumsum)(2n(-i+j))(.^(n)C_(i) + .^(n)C_(j))" "(2)` Adding (1) and (2), we have `2S = 2n underset(0leiltjlen)(sumsum)(.^(n)C_(i) +.^(n)C_(j))` or `S = n xx n2^(n) = n^(2)2^(n)` |
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| 54. |
Find the sum `(sumsum)_(0leiltjlen) jxx""^(n)C_(i)`. |
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Answer» `underset(0leiltjlen)(sumsum)j..^(n)C_(i)` `= underset(r=0)overset(n-1)sum.^(n)C_(r)[(r+1)+(r+2)+"...."+(n)]` `= underset(r=0)overset(n-1)sum.^(n)C_(r)[(r+1)+(r+2)+"...."(r+(n-r))]` `= underset(r=0)overset(n-1)sum.^(n)C_(r)(n-r)/(2)(r+1+n)` `= 1/2 underset(r=0)overset(n)sum.^(n)C_(r) (n(n+1)-r-r^(2))` ` = 1/2 [n(n+1)underset(r=0)overset(n)sum.^(n)C_(r)-underset(r=0)overset(n)sumr^(n)C_(r)-underset(r=0)overset(n)sumr^(2)..^(n)C_(r)]` `=1/2[n(n+1).2^(n)-n underset(r=0)overset(n)sum.^(n-1)C_(r-1)-n underset(r=0)overset(n)sumr..^(n-1)C_(r-1)]` `=1/2[n(n+1).2^(n)-n.2^(n-1)-n underset(r=0)overset(n)sum((n-1)..^(n-2)C_(r-2)+.^(n+1)C_(r-1))]` `= 1/2[n(2n+1).2^(n-1)-n(n-1).2^(n-2)-n.2^(n-1)]` `= n(3n+1).2^(n-3)` |
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| 55. |
Find the sum `(sumsum)_(0leiltjlen) ""^(n)C_(i)`. |
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Answer» `underset(0leiltjlen)(sumsum).^(n)C_(i) = ((underset(i=0)overset(n)sumunderset(j=0)overset(n)sum.^(n)C_(i))-underset(i=0)overset(n)sum.^(n)C_(i))/(2)` `= ((underset(i=0)overset(n)sum(n+1)^(n)C_(i))-underset(i=0)overset(n)sum.^(n)C_(i))/(2)` ` = ((n+1)2^(n)-2^(n))/(2)` `= n xx 2^(n-1)` |
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| 56. |
If `(1+x-2x^2)^(20)=a_0a_1x=a_2x^2+a_3x^3++a_(40)x^(40),`then find the value of `a_1+a_3+a_5++a_(39)dot` |
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Answer» `(1+x-2x^(2))^(20) = a_(0) + a_(1)x+a_(2)x^(2) + "….." +a_(40)x^(40)` Putting `x = 1`, we get `a_(0) + a_(1) + a_(2) + a_(3) +"……."+a_(40) = 0" "(1)` Putting, `x = - 1`, we get `a_(0) - a_(1) + a_(2) - a_(3) + "……" - a_(39) + a_(40) = 2^(20) " "(2)` Substracting (2) from (1), we get `2[a_(1)+a_(3)+"......"+a_(39)]=-2^(20)` or `a_(1)+a_(3)+"....."+a_(39)=-2^(19)` |
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| 57. |
If the sum of the coefficient in the expansion of `(alpha^2x^2-2alphax+1)^(51)`vanishes, then find the value of`alpha` |
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Answer» We have `(alpha^(2)x^(2)-3alphax+2)^(51)`. `:.` Sum of coefficients `= (alpha^(2)-3alpha+2)^(51)` Since sum of coefficient is zero, `alpha^(2) - 3alpha + 2 =0` `rArr (alpha+2)(alpha-1)=0` `rArr alpha = 1,2` |
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| 58. |
Prove that `.^(n)C_(1) - (1+1/2) .^(n)C_(2) + (1+1/2+1/3) .^(n)C_(3) + "…" + (-1)^(n-1) (1+1/2+1/3 + "…." + 1/n) .^(n)C_(n) = 1/n` |
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Answer» Let `S = .^(n)C_(1) - (1+1/2) .^(n)C_(2) + (1+1/2 + 1/3) .^(n)C_(3) + "….."` `+(-1)^(n-1)(1+1/2+1/3+"…."+1/n) .^(n)C_(n)`. rth term of the series is `T(r ) = (-1)^(r-1) . C_(r ) (1+1/2+1/3+"……"+1/r)`, where `C_(r) = .^(n)C_(r)` Let us consider a series whose general term is `T_(1)(r ) = (-1)^(r-1). C_(r)(1+x+x^(2)+"...."+x^(r-1))` `= (-1)^(r-1). C_(r) ((1-x^(r))/(1-x))` ` = ((-1)^(r-1)C_(r))/(1-x) + ((-1)^(r)x^(r)C_(r))/(1-x)` `rArr underset(r=1)overset(n)sum T_(1)(r ) = (1)/((x-1)) underset(r=1)overset(n)sum (-1)^(r) C_(r ) + (1)/((1+x))underset(r=1)overset(n)sum(-1)^(r). C_(r).x^(r )` `rArr underset(r=1)overset(n)sumT_(1)(r ) = (1)/((x-1)) (0-x) + (1)/((1-x)) ((1-x)^(n) - 1) = (1-x)^(n-1)` `= L` (say) Clearly `S = underset(0 )overset(1)intLdx = underset(0)overset(1)int(1-x)^(n-1) dx = 1/n` |
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| 59. |
Prove that ` sum_(n)^(r=0) ""^(n)C_(r)*3^(r)=4^(n).` |
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Answer» LHS=`.^(n)C_(0) xx 3^(0) + .^(n)C_(1) xx 3 + .^(n)C_(2) xx 3^(2)+...+^(n)C_(n)3^(n)` `=(1+3)^(n)=4^(n)= RHS " "` [by binomial expansion]. |
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| 60. |
Show that `(sqrt(2)+1)^6+(sqrt(2)-1)^6=198`A. 184B. 192C. 198D. 202 |
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Answer» Correct Answer - C N/a |
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| 61. |
Using binomial theorem, evaluate each of the following: (i)`(104)^(4)` (ii) `(98)^(4)` (iii)`(1.2)^(4)` |
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Answer» Correct Answer - (i) 104060401 (ii)92236816 (iii) 2.0736 (i) `(104)^(4)=(100+1)^(4)` `=.^(4)C_(0)(100)^(4)+.^(4)C_(1)(100)^(3) xx1+.^(4)C_(2)(100)^(2) xx1^(2)xx+.^(4)C_(3)(100) xx1^(3) +.^(4)C_(4) xx1^(4) `) `= 100000000 + 4000000 + 60000 + 400 +1=104060401` |
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| 62. |
Prove that `(2+sqrt(x))^(4)+(2-sqrt(x))^(4)= 2(16+24x+x^(2))`. |
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Answer» ` ( a + b ) ^ ( 4 ) + ( a - b ) ^ ( 4) = [.^ ( 4 ) C_(0) a^ ( 4 ) + .^ ( 4) C_(1) a^(3)b +.^(4)C_(2)a^(2)b ^ ( 2) + .^(4) C_(3)ab^(3) + .^(4)C_(4)b^(4)]` `+ [.^(4)C_(0) a^(4) + .^ (4)C_(1) a^(3) b + .^(4) C_(2)a^(2)b^(2) - .^(4)C_(3)ab^(3) + .^(4) C _(4)b^(4)]` `=2[.^(4) C _ (0) a^(4) + .^(4) C _(2) a^(2) b^(2) + .^(4) C _(4) b^(4) ]= 2 [a^(4) + 6a^(2) b^(2) + b^(4)].` Putting `a=2 and b= sqrt(x),` we get `(2+ sqrt(x) )^(4) + ( 2 - sqrt(x))^(4) = 2[2^(4) + 6 xx 2 ^(2) xx ( sqrt(x))^(2) + ( sqrt(x))^(4) ] = 2 ( 16+ 24x + x^(2)).` |
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| 63. |
The number of terms in the expansion of `(x^2+1+1/x^2)^n, n in N` , is:A. number of terms is `2n+1`B. constant term is `2^(n-1)`C. coefficient of `x^(2n-2)` is nD. coefficient of `x^(2)` in n |
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Answer» Correct Answer - A::C `(x^(2) + 1 + 1/(x^(2)))` `= .^(n)C_(0)+.^(n)C_(1)(x^(2) + 1/(x^(2)))+.^(n)C_(2)(x^(2) + 1/(x^(2)))^(2)+"......"+.^(n)C_(n)(x^(2)+1/(x^(2)))^(n)` This contains term having `x^(0), x^(2), x^(4), "…….."x^(2n), x^(-2n), x^(-4),"….",x^(-2n)` coefficient of constant term `= .^(n)C_(0) + (.^(n)C_(2))(2) + (.^(n)C_(4)) (.^(4)C_(2)) + (.^(n)C_(6)) (.^(6)C_(3)) + "......" ne 2^(n-1)`. Coefficient of `x^(2n-2)` is `.^(n)C_(n-1) = n` coefficient of `x^(2)` is `.^(n)C_(1) + (.^(n)C_(3))(.^(3)C_(1)) + (.^(n)C_(5))(.^(5)C_(2)) + "....." gt n` |
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| 64. |
The value of `.^(n)C_(1)+.^(n+1)C_(2)+.^(n+2)C_(3)+"….."+.^(n+m-1)C_(m)` is equal toA. `.^(m+n)C_(n) - 1`B. `.^(m+n)C_(n-1)`C. `.^(m)C_(1) + .^(m+1)C_(2) + .^(m+2)C_(3) + "…." + .^(m+n-1)C_(n)`D. `.^(m+n)C_(m) - 1` |
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Answer» Correct Answer - A::C::D `.^(n)C_(1)+.^(n+1)C_(2)+.^(n+2)C_(3)+"……"+.^(n+m-1)C_(m)` `= .^(n)C_(n-1)+.^(n+1)C_(n-1)+.^(n+2)C_(n-1)+"……"+.^(n+m-1)C_(n-1)` `=` Coefficient of `x^(n-1)` in `(1+x)^(n) [((1+x)^(m) -1)/((1+x) - 1)]` = Coefficient of `x^(n-1)` in `((1+x)^(m+n) -(1+x)^(n))/(x)` `=` Coefficient of `x^(n)` in `[(1+x)^(m+n) - (1+x)^(n)]` `= .^(m+n)C_(n) - 1` Similarly, we can prove `.^(m)C_(1)+.^(m+1)C_(2)+.^(m+2)C_(3)+"...."+.^(m+n-1)C_(n)=.^(m+n)C_(m)-1` |
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| 65. |
Prove that `sum_(r = 0)^(k) (-3)^(r -1) ""^(3n)C_(2 r - 1) = 0`, where `k = (3n)/2` and n is an even positive integer. |
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Answer» Since, n is an even positive integer, we can wirte `n =2, , m = 1,2,3….` Also , ` k = ( 3n)/2 = ( 3(2m))/2 = 3m` therefore `S = overset(3m)underset(r=1)sum(-3)^(r-1).^(6m)C_(2r-1)` i.e ` S = (-3)^(0).^(6m)C_(3) +... (-3)^(3m-1) . ^(6m)C_(3m-1)` From the binomial expanision , we write `(1+x)^(6m)=^(6m)C_(0)+(-3)^(6m)C_(3)+......+(-3)^(3m-1). ^(6m)C_(3m-1)` ` (1-x)^(6m) = ^(6m)C_(0) + ^(6m)C_(1)(-x)+^(6m)C_(2) (-x)^(2)+....+^(6m)C_(6m-1) (-x)^(6m-1)+ ^(6m)C_(6m) (-x)^(6m)` on subtracting Eq. (iii) from Eq. (ii) , we get ` (1+x)^(6m) - (1-x)^(6m)=2[.^(6m)C_(1)x+ ^(6m)C_(3)x^(3) + ^(6m)C_(5)x^(5x) +...+^(6m) C_(6m-1)x^(6m-1) ]` ` Rightarrow ((1+x)^(6m)-(1-x)^(6m))/(2x)=^(6m)C_(1)+^(6m)C_(3)x^(2)+ ^(6m)C_(5)x^(4) +.....+ ^(6m)C_(6m-1)x^(6m-2)` Let ` x^(2)=y` ` Rightarrow ((1+sqrty)^(6m)-(1-sqrty)^(6m))/(2sqrty)= ^(6m)C_(1)+^(6m)C_(3)y` ` + ^(6m)C_(5)y^(2)+...+^(6m)C_(6m-1)y^(3m-1)` For the required sum we have to put y=-3 in RHS. `S=((1+sqrt3)^(6m)-(1-sqrt(-3))^(6m))/(2sqrt(-3))` ` = ((1+isqrt3)^(6m)-(1-sqrt3)^(6m))/(2isqrt3)` Let `z= 1 sqrt3= r(cos theta + i sin theta)` ` Rightarrow r= |z|= sqrt(1 +3) =2` `and theta = pi //3` Now,` z^(6m)(cos 6m theta) + i sin 6m theta)` and `(overline(z))^(6m) = r^(6m) (cos 6m theta - i sin 6m theta)` ` Rightarrow z^(6m)- overline(z)^(6m) = r^(6m) (2 isin 6m theta)` Frow .Eq. (i) , ` S = (z^(6m)-overlinez^(6m))/(2isqrt3)= (r^(6m)(2isin6m theta))/(2isqrt3)` `= (2^(6m)sin 6m theta)/(sqrt3)` `0 "as" m inz, and theta= pi//3` |
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| 66. |
If `sum_(r=0)^(n) (n)/(""^(n)C_(0))= sum__(r=0)^(n) (n^(2)-3n+3)/(2.""^(n)C_(r))`, thenA. `n = 1`B. `n = 2`C. `n = 3`D. none of these |
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Answer» Correct Answer - A::C `underset(r=0)overset(n)sum(r)/(.^(n)C_(r)) = underset(r=0)overset(n)sum(n-r)/(.^(n)C_(n-r))= n/2underset(r=0)overset(n)sum(1)/(.^(n)C_(r))` `:. n/2 = (n^(2)-3n+3)/(2)` or `n = 1,3` |
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| 67. |
Prove that `sum_(r=1)^(k) (-3)^(r-1) ""^(3n)C_(2r-1) = 0` , where `k = 3n//2` and n is an even integer. |
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Answer» `S = underset(r=t)overset(k)sum (-3)^(r-1) .^(3n)C_(2r-1), k = (3n)/(2)` and n is even Let `k = (3(2m))/(2) = 3m` Then, `S = underset(r=1)overset(3m)sum(-3)^(r-1) xx .^(6m)C_(2r-1)` `= .^(6m)C_(1)-3.^(6m)C_(3)+3^(2).^(6m)C_(5)-"……"(-3)^(3m-1).^(6m)C_(6m-1)` `= 1/(sqrt(3))[sqrt(3).^(6m)C_(1)-(sqrt(3))^(3).^(6m)C_(3) + (sqrt(3))^(5).^(6m)C_(5)-"......"` ` + (-1)^(3m-1)(sqrt(3))^(6m-1).^(6m)C_(6m-1)]` There is an alternate sign series with odd binomial coefficients. Hence, we should replace x by `sqrt(3)i` in `(1+x)^(6m)`. Therefore, `(1+sqrt(3)i)^(6m) = .^(6m)C_(0)+.^(6m)C_(1)(sqrt(3)i)+.^(6m)C_(2)(sqrt(3)i)^(2)+.^(6m)C_(3)(sqrt(3)i)^(3)+"...."+.^(6m)C_(6m)(sqrt(3)i)^(6m)` `rArr sqrt(3) xx .^(6m)C_(1) -(sqrt(3))^(3).^(6m)C_(3)+(sqrt(3))^(5).^(6m)C_(5)+"...."` = Imaginary part in `(1+sqrt(3)i)^(6m)` `= "Im"[2^(6m)(1/2+(sqrt(3))/(2))^(6m)]` ` = "Im"[2^(6m)(cos2mpi + i sin 2m pi)]` ` = "Im" [2^(6m)] = 0` `rArr S = 0` |
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| 68. |
Prove that ` sum_(k=0)^(n) (-1)^(k).""^(3n)C_(k) = (-1)^(n). ""^(3n-1)C_(n)` |
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Answer» We have, `S = underset(k = 0)overset(n)sum(-1)^(k) . .^(3n)C_(k) = .^(3n)C_(0) - .^(3n)C_(1) + .^(3n)C_(2) + "……" + (-1)^(n)..^(3n)C_(n)` But `.^(3n)C_(0) = .^(3n-1)C_(0)` `-.^(3n)C_(1) = -.^(3n-1)C_(0) - .^(3n-1)C_(1)` `.^(3n)C_(2) = .^(3n-1)C_(1) .^(3n-1)C_(2)` `-.^(3n)C_(3) = -.^(3n-1)C_(2) - .^(3n-1)C_(3)` `{:("......"),("......"):}" "{:("......"),("......"):}` `(-1)^(n)..^(3n)C_(n) = (-1)^(n).^(3n-1)C_(n-1)+(-1)^(n).^(3n-1)C_(n)` On adding, we get `S = (-1)^(n).^(3n-1)C_(n)`. |
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| 69. |
If the second term of the expansion `[a^(1/(13))+a/(sqrt(a^(-1)))]^n`is `14 a^(5//2)`, then the value of `(^n C_3)/(^n C_2)`is. |
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Answer» Correct Answer - 4 `T_(2) = .^(n)C_(1) (a^(1//13))^(n-1).asqrt(a)= 14a^(5//2)` or `n.a^((n-1)/(13)) = 14a` or `n.a^((n-14)/(13)) = 14` or `(n-14)/(13) = 0` or `n = 14` `rArr (.^(14)C_(3))/(.^(14)C_(2)) = (14!)/(3!.11!) (2!12!)/(14!) = 12/3 = 4` |
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| 70. |
The sum of the series` ""^20 C_0-""^20 C_1+""^20C_2-""^20C_3+...-...+""^20C_10`is: |
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Answer» `(1+x)^n=^n C_0+^n C_1x+^n C_2x^2+.........^nC_nx^x` `if x=-1,x=20` `(1-1)^20=^20C_0-^20C_1+^20C_2-^20C_3.....^20C_19+^20C_20` `^nC_r=^nC_(n-r)` `0=2 ^20C_0-2^20C_1.....-^20C_9 2+2^20C_10 -^20C_10` `0=2(S)-^20C_10` `S=(^20C_10) /2` option`(2)` |
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| 71. |
The remainder left out when `8^(2n)""(62)^(2n+1)`is divided by 9 is(1) 0(2) 2(3) 7(4) 8 |
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Answer» `f(x) = 8^(2n) - (62)^(2n+1)` `= (9-1)^(2n) - (9 xx 7-1)^(2n+1)` `(1+x)^n = .^nC_0(1)^nx^0 + .^nC_1(x)^1 + ........ + .^nC_(n-1)x^(n-1) + .^nC_n x^n` remainder= `.^(2n)C_(2n) (-1)^(2n) - .^(2n+1)C_(2n+1)(-1)^(2n+1)` `= 1 xx 1 - 1xx(-1) = 1+1 = 2` Answer |
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| 72. |
The value of `(.^(21)C_(1) - .^(10)C_(1)) + (.^(21)C_(2) - .^(10)C_(2)) + (.^(21)C_(3) - .^(10)C_(3)) + (.^(21)C_(4) - .^(10)C_(4)) + … + (.^(21)C_(10) - .^(10)C_(10))` is |
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Answer» `(C(21,1) - C(10,1)) + (C(21,2) - C(10,2)) +...(C(21,10) - C(10,10))` Now, we know, `C(21,0) = C(10,0) = 1` So, we can write the given exprssion as , `(C(21,0) - C(10,0))+ (C(21,0) - C(10,1)) + (C(21,2) - C(10,2)) +...(C(21,10) - C(10,10))` `=[C(21,0) + C(21,1) +C(21,2) +...+ C(21,10) ] - [C(10,0)+C(10,1)+C(10,2)+...+C(10,10)]` `=1/2[C(21,0) + C(21,1) +C(21,2) +...+ C(21,21) ] - [C(10,0)+C(10,1)+C(10,2)+...+C(10,10)]` `= 1/2(2^21) - 2^10` `=2^20-2^10.` So, option `(2)` is the correct option. |
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| 73. |
The value of `(.^(21)C_(1) - .^(10)C_(1)) + (.^(21)C_(2) - .^(10)C_(2)) + (.^(21)C_(3) - .^(10)C_(3)) + (.^(21)C_(4) - .^(10)C_(4)) + … + (.^(21)C_(10) - .^(10)C_(10))` isA. `2^(20) - 2^(10)`B. `2^(21) - 2^(11)`C. `2^(21) - 2^(10)`D. `2^(20) - 2^(9)` |
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Answer» Correct Answer - A We have `(.^(21)C_(1)+.^(21)C_(2)"....."+.^(21)C_(10))-(.^(10)C_(1)+.^(10)C_(2)"....".^(10)C_(10))` `= 1/2[.^(21)C_(0)+.^(21)C_(1)+.^(21)C_(2)+"..."+.^(21)C_(21)]-2^(10)` `= 2^(20) - 2^(10)` |
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| 74. |
The value of `(.^(21)C_(1) - .^(10)C_(1)) + (.^(21)C_(2) - .^(10)C_(2)) + (.^(21)C_(3) - .^(10)C_(3)) + (.^(21)C_(4) - .^(10)C_(4)) + … + (.^(21)C_(10) - .^(10)C_(10))` is(A) `2^(21) - 2^(11)`(B) `2^(21) - 2^(10)`(C) `2^(20) - 2^(9)`(D) `2^(20) - 2^(10)`A. `2^(21) - 2^(11)`B. `2^(21) - 2^(10)`C. `2^(20) - 2^(9)`D. `2^(20) - 2^(10)` |
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Answer» `(.^(21)C_(1) - .^(10)C_(1)) + (.^(21)C_(2) - .^(10)C_(2)) + (.^(21)C_(3) - .^(10)C_(3)) + …. + (.^(21)C_(10) - .^(10)C_(10))` `= (.^(21)C_(1) + .^(21)C_(2) + … + .^(21)C_(10)) - (.^(10)C_(1) + .^(10)C_(2) + ….+ .^(10)C_(10))` `= (1)/(2) (.^(21)C_(1) + .^(21)C_(2) + .... + .^(21)C_(20)) - (2^(10) - 1)` `= (1)/(2) (.^(21)C_(1) + .^(21)C_(2) + .... + .^(21)C_(21) - 1) - (2^(10) - 1)` `= (1)/(2) (2^(21) - 2) - (2^(10) - 1) = 2^(20) - 1 - 2^(10) + 1 = 2^(20) - 2^(10)` |
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| 75. |
If the number of terms in the expansion of `(1-2/x+4/(x^2))^n , x!=0,`is 28, then the sum of the coefficients ofall the terms in this expansion, is :(1) 64(2) 2187(3) 243(4) 729A. 64B. 2187C. 243D. 729 |
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Answer» Clearly, number of terms in the expansion of `(1 - (2)/(x) + (4)/(x^(2)))^(n)` is `((n + 2)(n + 1))/(2)` or `.^(n + 2)C_(2)` assuming `(1)/(x)` and `(1)/(x^(2))` distinct] `:.((n+2) (n + 1))/(2) = 28` `implies (n + 2) (n + 1) = 56 = (6 + 1) (6 + 1) implies n = 6` Hence, sum of coefficients `= (1 - 2 + 4)^(6) = 3^(6) = 729` Note As `(1)/(x)` and `(1)/(x^(2))` are functions of same variables therefore number of dissimilar terms will be `2n + 1`, i.e., odd, which is not possible, Hence it contains error. |
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| 76. |
If the number of terms in the expansion of `(1-2/x+4/(x^2))^n , x!=0,`is 28, then the sum of the coefficients ofall the terms in this expansion, is :(1) 64(2) 2187(3) 243(4) 729A. 2187B. 243C. 729D. 64 |
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Answer» Correct Answer - C Theroectically the number of terms are `2n+1`(i.e, odd) But given that number of term is `28`. So considering number of term `= .^(n+2)C_(2) = 28`. (Here we are ignoring clubbing of terms) `:. N = 6` `:.` Sum of coefficient `= 3^(n) = 3^(6) = 729` (Putting `x = 1`) |
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| 77. |
The sum of the co-efficients of all odd degree terms in the expansion of (x+sqrt(x^3-1))^5+(x-(sqrt(x^3-1))^5`, (x gt 1)`A. 2B. -1C. 0D. 1 |
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Answer» Correct Answer - A `(x+sqrt(x^(3)-1))^(5) + (x-sqrt(x^(3)-1))^(5)` `= 2[.^(5)C_(0)x^(5)+.^(5)C_(2)x^(3)(sqrt(x^(3)-1))^(2) + .^(5)C_(4)x(sqrt(x^(3)-1))^(4)]` `=2[x^(5)+10x^(3)(x^(3)-1)+5x(x^(3)-1)^(2)]` `= 2[x^(5)+10x^(6)-10x^(3)+5x^(7)-10x^(4)+5x]` `:.` Sum of the coeffcient of odd degree terms `= 2(1-0+5+5) = 2` |
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| 78. |
If m and n are positive integers, then prove that the coefficients of `x^(m) " and " x^(n)` are equal in the expansion of `(1+x)^(m+n)` |
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Answer» In the binomial expansion of `(1+x)^(m+n)` , the general term is given by ` T_(r+1) = ^(m+n)C_(r) xxx^(r)`. `:."coefficient of "x^(m) " in the expansion of "(1+x)^(m+n)` `.^(m+n)C_(m)=((m+n)!)/({(m+n-m)!}xx (m!))=((m+n)!)/((m!) xx(n !)).` coefficient of `x^(n)" in the expansion of "(1+x)^(m+n)` `=.^(m+n)C_(n)=((m+n)!)/({(m+n-n)!} xx(n!))+((m=n)!)/((m!) xx(n!)).` Hence, the coefficents fo `x^(m) and x^(n)` are equal. |
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| 79. |
Find the coefficent of`x^(6)y^(3)` in the expansion of `(x+2y)^(9)`. |
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Answer» The general term in the expansion of `(x+2y)^(9)` is given by `T_(r+1)=.^(9)C_(r) xx x^((9-r)) xx(2y)^(r)` `rArr T_(r+1)=.^(9)C_(r) xx2^(r) xxx^((9-r)) xxy^(r)." "`...(i) We have to find the coeffcient of `x^(6)y^(3)`. Putting r=3 in (i), we get `T_(3+1)=.^(9)C_(3) xx 2^(3) xx (x^(6)y^(3))`. `:." coefficient of "x^(6) y^(3)` in the given expansion `=(.^(9)C_(3) xx 2^(3))=((9 xx8xx7)/(3xx2xx1) xx8) =672`. Hence, the coefficient of `x^(6)y^(3)` in the given expansion is 672. |
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| 80. |
Find the coefficient of `x^(4)` in the expansion of `(1+x+x^(2)+x^(3))^(11)`. |
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Answer» We have `(1+x+x^(2) + x^(3))^(11) = {(1+x)+x^(2)(1+x)}^(11)={(1+x)(1+x^(2))}^(11)` `=(1+x)^(11) xx(1+x^(2))^(11)` `=(.^(11)C_(0)x^(0)+.^(11)C_(1)x_ +.^(11)C_(2)x^(2)+.^(11)C_(3)x^(3)+.^(11)C_(4)x^(4)+...)` ` xx[.^(11)C_(0)(x^(2))^(0) +.^(11)C_(1)(x^(2)) + .^(11)C_(2)(x^(2))^(2)+...]` `(1+11x+55x^(2)+165x^(3) + 330x^(4)+...)` `xx(1+11x^(2)+55x^(4)+...).` `:. "coefficient of " x^(4) " in the given expansion "` `=(1 xx55) + ( 55 xx11) +(330 xx1)` `=(55+605 + 330)=990`. Hence, the coefficient of `x^(4)` in the given expansion is 990. |
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| 81. |
Find the coefficent of ` x^(4)` in the product `(1+2x)^(4) xx (2-x)^(5).` |
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Answer» Using the binomial expansion, we get `(1+2x)^(4) = .^(4) C _(0) + .^(4) C _(1) ( 2x) + .^(4) C _(2) (2x)^(2) + .^(4) C _(3) ( 2x) ^ (3) + .^(4) C _(4) ( 2x) ^ (4)` `=1 + 8x + 24x^(2) + 32x^(3) + 16x^(4)` and `( 2-x) ^ (5) = 2^(5) - . ^ (5) C _(1) (2^(4))x + .^(5) C _ (2)(2^(3))x^(2) - .^(5) C _(3) ( 2^(2))x^(3)` `=.^(5) C _(4) ( 2) x^(4) - .^(5) C _ (5) x ^ (5)` `= 32 - 80x + 80x^(2) - 40x^(3) + 10x^(4) - x^(5)` ` :. (1+2x)^(4) xx (2-x)^(5)` ` =(1+ 8x + 24x^(2) + 32x^(3) + 16x^(4))` `xx ( 32 - 80x + 80x^(2) - 40x^(3) + 10x^(4) - x^(5)).` Sum of the terms containing `x^(4)` in the given product `(1 xx 10x^(4)) + 8x (-40x^(3)) + (24x^(2)) (80x^(2))` ` ( 32 x^(3) ) (-80x) + ( 16x^(4) xx 32 ) ` `= 10^(x^(4)) - 320 x^(4) + 1920x^(4) - 2560x^(4) + 512 x^(2) ` `= (10- 320 + 1920 - 2560 + 512) x^(4) = - 438x^(4)`. Hence , the coefficient of `x^(4)` in the given product is `-438`. |
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| 82. |
If P be the sum of all odd terms and Q that of all even terms in the expansion of `(x+a)^(n)`, prove that |
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Answer» By the binomial expansion, we have `( x+ a) ^(n) = . ^(n) C _ (0) x^(n) + .^(n) C _ ( 1) x^(n-1) + .^(n) C _(2)x ^ ( n-2) + .^ (n) C_(3)x^ (n-3) + . ^ (n)C_(4) x^ ( n-4) + ... + .^(n) C _(n) a^(n).` `(T_(1) + T _(2) + T _ (3) + T _ (4) + T _( 5) + ... + T _ (n ) + T _ ( n+ 1 ) )` `( T _ ( 1 ) + T _ (3) + T _ ( 5 )...) + (T _ ( 2) + T _ ( 4 ) + ... + T_ ( 6 ) + ...)` `= P + Q." "` ...(I) And `( x - a) ^ (n) + .^ ( n) C _(0) x ^ (n ) - . ^ (n ) C _ ( 1 ) x ^ ( n-1) a + . ^ ( n) C _ ( 2 ) x ^ ( n-2) a ^ ( 2 ) - . ^ ( n) C _ ( 3 ) x ^ ( n-3) a^ ( 3 ) ` ` ...+(-1)^(n) .^(n)C_(n)a^(n)` `= ( T _ ( 1 ) - T _ ( 2 ) + T _ ( 3 ) - T _ ( 4 ) + ...)` `= ( T _ ( 1 ) + T _ ( 3 ) + T _ ( 5 ) + ...) - ( T _ ( 2 ) + T _ ( 4 ) + T _ ( 6 ) ...)` `= P- Q." " ... ( II)` (i ) On multiplying ( I ) and ( I I ) , we get `(x^ ( a) - a ^ ( 2 ) ) ^ ( n) = ( p ^ ( 2 ) _ Q ^ ( 2 ) ) .` On squaring ( I ) and (I I ) and subtracting , we get `{( x + a) ^ ( 2n) - ( x - a ) ^ ( 2n) } = { ( P + Q ) ^ ( 2 ) - ( P - Q ) ^ ( 2 ) } = 4 P Q .` ( iii) On squaring ( I ) and ( I I) and adding , we get `{ ( x + a ) ^ ( 2n) + ( x - a ) ^ ( 2n) } = { (P + Q ) ^ ( 2 ) + ( P - Q ) ^ ( 2 ) } = 2 ( P ^ ( 2 ) + Q ^ ( 2 ) ) .` |
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| 83. |
Largest real value for x such that `sum_(k=0)^(4) ((3^(4-k))/((4-k)!))((x^(k))/(k!))= 32/3` |
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Answer» Correct Answer - 1 `underset(k=0)overset(4)sum((3^(4-k))/((4-k)!))((x^(k))/(k!))` `=underset(k=0)overset(4)sum((4!)/((4-k)!k!)3^(4-k).x^(4).(1)/(4!))` `= underset(k=0)overset(4)sum(.^(4)C_(k).3^(4-k).x^(4))/(04!)` `= ((3+x)^(4))/(4!)` According to the question , `((3+x)^(4))/(4!) = 32/3` or `(3+x)^(4) = 256` or `x + 3 = 256` or `x = 1` |
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| 84. |
Using binomial theorem, find the value of`(103)^(4)`. |
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Answer» By the binomial expansion, we have `(103)^(4) = (100+3)^(4)` ` .^(4)C_(0)*(100)^(4)+.^(4) C_(1) xx(100)^(3)xx3+ .^(4)C_(2) xx(100)^(2) xx3^(2)+.^(4)C_(3) xx100 xx3^(3)+.^(4)C_(4) xx3^(4)` `(100)^(4) + 12 xx (100)^(3)+ 54 xx(100)^(2) + 108 xx 100 + 81` `=100000000 + 12 xx 1000000 + 540000 + 10800 + 81` `= 100000000 +12000000 + 540000 + 10800 + 81` `=112550881.` Hence, `(103)^(4) = 112550881.` |
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| 85. |
The coefficient of `x^(103)` in `(1+x+x^(2) +x^(3)+x^(4))^(199)(x-1)^(201)` is `"___"`. |
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Answer» `E = (1+x+x^(2)+x^(3)+x^(4))^(199) (x-1)^(201)` `= ((1-x^(5))/(1-x))^(199)(x-1)^(201)` `= - (1-x)^(2)(1-x^(5))^(199)` `= - 1(1-2x+x^(2))(1-.^(199)C_(1)x^(5)+.^(199)C_(2)x^(10)-"…..")` `:.` Coefficient `x^(103)` in `E = 0` |
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| 86. |
The total number of different terms in the product `(.^(101)C_(0) - .^(101)C_(1)x+.^(101)C_(2)x^(2)-"….."-.^(101)C_(101)x^(101))(1+x+x^(2)+"…."+x^(100))^(101)` is `"____"`. |
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Answer» Correct Answer - 102 `(.^(101)C_(0).^(101)C_(1)x+.^(101)C_(2)x^(2)-"...."-.^(101)C_(101)x^(101))(1+x+x^(2)+"...."+x^(100))^(101)` `= (1-x)^(101).((1-x^(101))/(1-x))^(101)` `= (1-x^(101))^(101)` `:.` Number of different terms `= 102` |
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| 87. |
The value of `sum_(r=0)^(3) ""^(8)C_(r)(""^(5)C_(r+1)-""^(4)C_(r))` is `"_____"`. |
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Answer» Correct Answer - 220 `underset(r=0)overset(3)sum.^(8)(C_(r+1)-.^(4)C_(r))=underset(r=0)overset(3)sum.^(8)C_(r).^(4)C_(r+1)` `= .^(8)C_(0) xx.^(4)C_(1)+.^(8)C_(1)xx.^(4)C_(2)+.^(8)C_(2)xx.^(4)C_(3)+.^(8)C_(3)xx.^(4)C_(4)` `=` coefficient of `x^(3)` in `(1+x)^(4)(1+x)^(8)` `=` coeficient of `x^(3)` in `(1+x)^(12)` `=.^(12)C_(3) = 220` |
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| 88. |
the value of `x`, for which the 6th term in the expansions of`[2^log_2sqrt(9^((x-1)+7))+1/(2^(1/5)(log)_2(3^(r-1)+1))]i s84`, is equal toa. 4 b. 3 c. `2`d. `1`A. 4B. 3C. 2D. 1 |
| Answer» Correct Answer - C | |
| 89. |
Find the term independent of x in the expansion of `(2^(x) + 2^(-x)+log_(e)e^(x+2)))^(30)`. |
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Answer» We have `(2^(x) + 2^(-x)+log_(e)e^(x+2))^(30)` `= (2^(x)+2^(-x)+x+2)^(30)` ` = ((2x^(x//2) + 2^(-x//2))^(2) + x )^(30)` Therefore, the term independent of x occurs in `.^(30)C_(0)(2^(x//2) + 2^(-x//2))^(60)` which is `.^(30)C_(0) xx .^(60)C_(30)` or `.^(60)C_(30)`. |
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| 90. |
If the third in the expansion of `[x + x^(logx)]^(6)` is ` 10^(6)` , then x (x>1) may beA. 1B. 10C. `10^(-5//2)`D. 102 |
| Answer» Correct Answer - B | |
| 91. |
The value of `x`in the expression `(x+x^((log)_(10)))^5`if third term in the expansion is 10,00,000 is/area. 10 b. 100 c. `10^(-5//2)`d. `10^(-3//2)`A. 10B. 100C. `10^(-5//2)`D. `10^(-3//2)` |
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Answer» Correct Answer - A::C Inclusion of log x implies `x gt 0` Now, `3^(nd)` term in expension is `T_(2+1) = .^(5)C_(2)x^(5-2)(x^(log_(10)x))^(2) = 10000000` (given) or `x^(3+2log_(10)x) = 10^(5)` Taking logarithm of both sides, we get `(3+2log_(10)x)log_(10)x = 5` or `2y^(2) + 3 y - 5 = 0`, where `log_(10) x = y` or `(y-1) (2y+5) = 0 y = 1` or `-5//2` or `log_(10)x=1` or `-5//2` `:. x = 10^(1) = 10` or `10^(-5//2)` |
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| 92. |
If the `4^(th)` term of `{sqrt(x^((1)/(1+log_(10)x)))+root(12)(x)}^(6)` is equal to `200`, `x gt 1`and the logarithm is common logarithm, then`x` is not divisible byA. `2`B. `5`C. `10`D. `4` |
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Answer» Correct Answer - D `(d)` Given expression is `{sqrt((1)/(x^(1+log_(10)x)))+root12x}^(6)={x^((1)/(2))((1)/(1+log_(10x)))+x^((1)/(12))}^(6)` `T_(4)=200` `implies^(6)C_(3)*x^((1)/(2)((1)/(1+log_(10)x))^(6-3))xx(x^((1)/(12)))^(3)=200` `impliesx^((3)/(2)(1)/(1+log_(10)x)+(1)/(4))=10` `=(3)/(2(1+log_(10)x))+(1)/(4)=log_(x)10` `implies(3)/(2(1+log_(10)x))+(1)/(4)=(1)/(log_(10)x)` Put `log_(10)x=y` `implies(3)/(2(1+y))+(1)/(4)=(1)/(y)` `impliesy^(2)+3y-4=0` `impliesy=-4`, `y=1` `implieslog_(10)x=-4`, `log_(10)x=1` `impliesx=10` (as `x gt 1`) |
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| 93. |
Show that coefficient of `x^(-3)` in the expansion of `(x-1/x)^(11) is -330`. |
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Answer» `T_(r+1) = (-1)^(r) * . ^(11)C_(r)x^((11-1)) * (1/x)^(r) = (-1)^(r) *.^(11)C_(r)x^((11-2r)).` Now `11- 2r = - 3 rArr 2r = 14 rArr r = 7.` `T_(8) = (-1)^(7)*.^(11)C_(7) x^(-3) = -.^(11) C _(4)x^(-3) = - ( 11 xx 10 xx 9 xx 8)/(4 xx 3 xx 2 xx 1) x^(-3) = -330x^(-3).` |
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| 94. |
If the coefficents of `x^(2) and x^(3)` in the expansion of `(3+px)^(9)` are the same then prove that `p= 9/7`. |
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Answer» `T_(r+1)= .^(9)C_(r)3^((9-1)) xx(px)^(r) = .^(9)C_(r)3^((9-r))* p ^(r)x^(r).` `" Coeff. of " x^(2) = " coeff.of " x^(3) rArr .^(9) C_(2) xx 3^(7) xx p^(2) = .^(9) C _(3) xx 3^(6) xx p^(3).` |
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| 95. |
No. of terms in the expansion of `(1+2x)^(9) +(1-2x)^(9)` is :A. 10B. 9C. 7D. 5 |
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Answer» Correct Answer - D N/a |
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| 96. |
Find the middle term in the expansion of : ` (x-1/x)^(10)`A. 126B. -126C. -252D. 252 |
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Answer» Correct Answer - C N/a |
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| 97. |
The coefficient `x^5` in the expansion of `(2 - x + 3x^2)^6` isA. `-5051`B. 4632C. `-4631`D. none of these |
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Answer» Correct Answer - A N/a |
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| 98. |
Prove that `C_(1)^(2) - 2. C_(2)^(2) + 3. C_(3)^(2) - …. -2n . C_(2n)^(2) = (-1)^(n). C_(n)` |
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Answer» We know that , `(1 + x )^(2n) = C_(0) + C_(1)x + C_(2) x^(2) + …+ C_(2n) x^(2n)` On differntiating both sides w.r.t.x, we get ` 2n (1 + x)^(2n-1) = C_(1) + 2.C_(2) x + 3. C_(3) x^(3) + …+ 2nC_(2n x^(2n-1) ` …(i) and ` (1 - (1)/(x))^(2n) = C_(0) - C_(1).(1)/(x) + C_(2) .(1)/(x^(2)) C_(3) . (1)/(x^(3)) + ...+ C_(2n). (1)/(x^(2n))` ...(ii) On multiplying Eqs. (i) and (ii) , we get ` 2n (1 + x)^(2n-1) (1 - (1)/(x))^(2n)` = `[C_(1) + 2*C_(2)x + 3. C_(3) x^(2) +...+ 2n*C_(2n)x^(2n-1)] xx[C_(0) - C_(1) ((1)/(x)) + C_(2) ((1)/(x^(3))) -...+C_(2n) ((1)/(x^(2n)))]` Coefficent of ` ((1)/(x))` on the LHS = Coefficient of ` (1)/(x) `in 2n `((1)/(x^(2n)))(1 + x)^(2n - 1) (x - 1)^(2n)` Coefficient if ` x^(2n - 1)` in 2n ` (1 - x^(2))^(2-1) (1 - x)` ` 2n (-1)^(n-1) . (2n -1) C_(n-1) (-1)` ` = (-1)^(n) (2n) ((2n -1)!)/((n-1)!n!) = (-1)^(n)n((2n)!)/((n!^(2)))n` ` = - (-1)^(n)n . C_(n)` ...(iii) Again , the coefficient of `((1)/(x))` on the RHS ` = - (C_(1)^(2) - 2 *C_(2)^(2) + 3*C_(3)^(2)-...-2nC_(2n)^(2))` ...(iv) From Eqs. (iii) and (iv), `- C_(1)^(2) - 2 *C_(2)^(2) + 3*C_(3)^(2)-...-2nC_(2n)^(2) = (-1)^(n) n.C_(n)`. |
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| 99. |
Prove that `(.^(2n)C_(0))^(2) - (.^(2n)C_(1))^(2) + (.^(2n)C_(2))^(2) - …. + (.^(2n)C_(2n))^(2) = (-1)^(n) .^(2n)C_(n)` |
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Answer» `(1+x)^2n(1-(1)/(x))^2n` `=(.^2nC_0 -(.^2n C_1)x+(.^2nC_2)x62+.....+(.^2nC_2n)x^2n]` `xx[.^2nC_0 -(.^2nC_1)(1)/(X)+(.^2n C_2)(1)/(x^2)+.....+(.^2n C_2n)(1)/(x^2n)]` Independent terms of x on RHS `=(.^2n C_0)^2-(.^2n C_1)^2+(.^2nC_2)^2 -......+(.^2n C_2n)^2`. LHS `=(1+x)^2n((x-1)/(x))^2n =(1)/(x^2n)(1-x^2)^2n`. Independent term of x on the LHS `=(-1)^n .^2n C_n`. |
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| 100. |
`^404C_4-^303C_4.^4C_1+^202C_4.^4C_2-^101C_4.^4C_3=`A. `(401)^(4)`B. `(101)^(4)`C. `0`D. `(201)^(4)` |
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Answer» Correct Answer - B The given expression is the coefficient of `x^(4)` in `.^(4)C_(0)(1+x)^(404)-.^(4)C_(1)(1+x)^(303)+.^(4)C_(2)(1+x)^(202)-.^(4)C_(3)(1+x)^(101)+.^(4)C_(4)` = Coefficient of`x^(4)` in `[(1+x)^(101)-1]^(4)` `=` Coefficient of `x^(4)` in `(.^(101)C_(1)x+.^(101)C_(2)x^(2)+".....")^(4)` `= (101)^(4)` |
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