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InterviewSolution
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.
| 1. |
In the expansion of `(5^(1//2)+7^(1//8))^(1024),`the number of integral terms is`128`b. `129`c. `130`d. `131`A. 128B. 129C. 130D. 131 |
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Answer» Correct Answer - b The general term `T_(r + 1)` in the expansion of `(5^(1//2) + 7^(1//8)) ^(1024)` is given by `T_(r+ 1) = ""^(1024)C_(r) (5^(1//2))^(1024 - r)(7 ^(1//8))^(r)` `rArr T_(r + 1) = ""^(1024)C_(r) 5 ^(512 - r//2) 7^(r//8)` `rArr T_(r +1) = (1024C_(r) 5^(512 - r)) (5 ^(r//2) 7^(r//8))` `rArr T_(r+ 1) = { ""^(1024)C_(r) 5^(512 - r)} (5^(4) xx 7 )^(7)^(r //8)` Clearly, `T_(r +1)` will be an integer , iff ` r//8` is an integer such that 0 `le r le 1024` `rArr r` is a multiple of 8 lying satisfying ` 0 le r le 1024` `rArr ` r = 0 , 8 , 16 ..., 1024 `rArr ` r can assume 129 values Hence, these are 129 integral terms in the expansions of `(5^(1//2) + 7 ^(1//8))^(1024)` |
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| 2. |
For natural numbers m, n if `(1-y)^(m)(1+y)^(n) = 1+a_(1)y+a_(2)y^(2) + "……."` and `a_(1) = a_(2) = 10`, then `(m,n)` is :A. (20, 45)B. (35, 20)C. `(45, 35)`D. (35, 45) |
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Answer» Correct Answer - d We have, `(1 - y)^(m) (1 + y)^(n)` `(""^(m)C_(0) - ""^(m)C_(1)y ^(2) -...)xx(""^(n)C_(0) + ""^(n)C_(1) y + ""^(n)C_(2) y^(2) +...)` =` 1 + (n+ m) y + {(m(m-1) + n (n-1))/(2) -mn }y^(2) +…` `therefore a_(1) = n - m and a_(2) = (m (m -1) + n(n-1))/(2) - mn` `rArr n-m = 10 and (m^(2) + n^(2) - (m+n ) - 2mn)/(2) = 100 [ because a_(1) = a_(2) = 10]` `rArr n-m = -10 and (m - n)^(2) - (m +n) = 20` `rArr m - n = - 10 and m + n = 80` `rArr m = 35 and n = 45` . |
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| 3. |
If the expansion in power of x of the function `(1)/(( 1 - ax)(1 - bx)) is a_(0) + a_(1) x + a_(2) x^(2) + a_(3) a^(3) + …, ` then `a_(n)` isA. `(b^(n) - a^(n))/(b-a)`B. `(a^(n) - b^(n))/(b-a)`C. `(a^(n+1) - b^(n+1))/(b-a)`D. `(b^(n+1) - a^(n+1))/(b-a)` |
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Answer» Correct Answer - d We have, `(1)/((1 - ax)(1 - bx) )=(1)/(a -b) {(a)/(1 - ax) (b)/(1 - bx)}` `rArr (1)/((1 - ax)(1 - bx) )=(1)/(a -b) {a(1 - ax)^(-1) - b(1 - bx)^(-1)}` ltbegt `rArr (1)/((1 - ax)(1 - bx) )=(1)/(a -b) {asum_(r=0)^(infty) ( ax)^(1) - b(1 - bx)^(-1)}` `rArr (1)/((1 - ax)(1 - bx) )=(1)/(a -b) {a(sum_(r=0)^(infty) a^(r)x^(r)) - b(sum_(r= 0)^(infty) b^(r) x^(r))^()}` `therefore a_(n)` = Coefficient of `x^(n)` in` (1)/((1 - ax) (1 - bx))` `rArr a_(n) = (1_)/(a - b) { axx a^(n) - b xxb^(n)} = ((a^(n +1) - b^(n +1))/(a -b) `. |
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| 4. |
The value of `(30 0)(30 10)-(30 1)(30 11)+(30 2)(30 12)++(30 20)(30 30)=``^60 C 20`b. `^30 C 10`c. `^60 C 30`d. `^40 C 30`A. `""^(30)C_(11)`B. `""^(60)C_(10)`C. `""^(30)C_(10)`D. `""^(65)C_(55)` |
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Answer» Correct Answer - c We have, `""^(30)C_(0)xx""^(30)C_(10)- ""^(30)C_(1)xx""^(30)C_(11)+...+ ""^(30)C_(10)xx""^(30)C_(0)` = Coefficient of `x^(20)` in `{(1 + x )^(30) (1 - x)^(30)}` = Coefficient of `x^(20)` in ` (1 - x^(2))^(30)` ` = ""^(30)C_(10) (-1)^(10) = ""^(30)C_(10)` |
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| 5. |
Which is larger : `(99^(50)+100^(50))` or `(101)^(50)`.A. `99^(50) + 100^(50)`B. both are equalC. ` (101)^(50)`D. none of these |
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Answer» Correct Answer - c We have , ` 99^(50) + 100 ^(50) - 101^(50)` ` = (100 - 1 )^(50) - (1 - 100)^(50) + 1000^(50)` ` = - {(1 +100)^(50) - (1 - 100)^(50)} + 100^(50)` ` = - 2 {""^(50)C_(1)xx100 +""^(50)C_(3)xx100^(3) + ... + ""^(50)C_(49)xx100^(49)} + 100^(50)` ` = - 2 {""^(50)C_(1)xx100 +""^(50)C_(3)xx100^(3) + ... + ""^(50)C_(47)xx100^(49)} lt 0` ` therefore 99^(50) + 100^(50) - 101 ^(50) lt 0 rArr 00^(50) + 100^(50) lt 101^(50)`. |
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| 6. |
If w is a non-real cube root of unity, x is a real number and `n in N` such that first three terms in the binomial expansion of `(w + x)^n` are `1,12bar w and 69 w` respectively, thenA. ` n = 36, x = 1`B. n = 12, x=2C. ` n = 24, x = (1)/(2)`D. ` n = 18, x = (1)/(3)` |
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Answer» We have , `( omega + x)^(n) = ""^(n)C_(0) omega^(n) x^(0) + ""^(n)C_(1) omega^(n-1) x + ""^(n)C_(2) omega^(n-2) x^(2) +…+ ""^(n)C_(n) x^(n)` `therefore omega^(n) = 1, n omega^(n-1) x= 12bar(omega) and ""^(n)C_(2) omega^(n -2) x^(2) = 69 omega` Now, `omega^(n) = 1 rArr n= 3 m, m in` N `therefore n omega ^(n-1) x = 12 bar(omega) and ""^(n)C_(2) omega^(n-2) x^(2) = 69 omega` `rArr 3m x = 12 and (3m (3m -1))/(2) x^(2) = 69` `rArr mx = 4 and 3m^(2) x^(2) - (mx) x = 46` `rArr mx = 4 and 3xx 16 - 4x = 46` `rArr x = (1)/(2) and m=8` `rArr x= (1)/(2) and n=24` [because n=3m]` |
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| 7. |
The coefficient of `x^9` in the expansion of `(1 + x) (1 +x^2)(1 +x^3)...(1+x^(100))`isA. 2B. 6C. 9D. 8 |
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Answer» Correct Answer - d We observe that `x^(9)` can be formed in the following 8 ways : `x^(9+0), x^(1+8), x^(2 +7), x ^(3 +6), x^(4 +5), x^(1 + 3 + 5), x^(1 + 2 + 6), x ^(2 + 3 + 4)` and the coefficients in each case is 1. `therefore ` Coefficients of `x^(9) = underset(("8 - times"))(1 + 1 + ... + 1 = 8)` ALITER The coefficients of `x^(9)` is the number of ways in which the sum of the power of x is 9 as listed below. `x^(9) , x^(3) xxx^(8), x^(7), x^(3)xxx^(6), x^(4) xxx^(5), x^(1) xxx^(2)xxx^(6), x^(1) xxx^(3)xxx^(5),x^(2) xx^(3)xxx^(4)`. Clearly, these are 8 ways. |
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| 8. |
The power of x which has the greatest coefficient inthe expansion of `(1+x/2)^10`A. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - b Let the coefficient of `(r +1)^(th)` term be the greatest coefficient in the expansion of `(1 + (x)/(2))^(10)`. Then, `((T_(r+1))/(T_(r)))_(x=1) gt 1 and ((T_(r+2))/(T_(r+1)))_(x=1)lt1` `rArr (""^(10)C_(r)((1)/(2))^(r))/(""^(10)C_(r-1)((1)/(2))^(r-1)) gt 1 and (""^(10)C_(r+1)((1)/(2))^(r+1))/(""^(10)C_(r)((1)/(2))^(r))lt1` `rArr (11 -r)/(2r) gt 1 and (10 -r)/(2(r +1)) lt 1` `rArr 8 lt 3r lt 11` `rArr r = 3` Now, `T_(r+1) = ""^(10)C_(r) ((1)/(2))^(r) rArr T_(4) = ""^(10)C_(3) ((1)/(2))^(3) x^(3)` Hence, the power of x having the greatest coefficient in 3. |
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| 9. |
The coefficient of `x^2 y^5 z^3` in the expansion of `(2x + y + 3z)^10` isA. `(10!)/(2!3!5!)`B. `(10!)/(2!3!5!)xx2^(2) xx3^(3)`C. `(10!)/(2!3!5!)xx2^(3) xx3^(2)`D. `10! xx2^(2) xx3^(3)` |
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Answer» Correct Answer - b We have, `(2x +y +3z)^(10) = sum_(r_(1)+r_(2)+r_(3) = 10) (10!)/(r_(1)!r_(2)!r_(3)!)(2x)^(r_(1)) (y)^(r_(2)) (3z)^(r_(3))` The general term in the above expansion is (10!)/(r_(1)!r_(2)!r_(3)!)(2x)^(r_(1)) (y)^(r_(2)) (3z)^(r_(3))=(10!)/(r_(1)!r_(2)!r_(3)!)2^(r_(1))3^(r_(3))x^(r_(1))y^(r_(2)) z^(r_(3)` For the coefficient of `x^(2) `y^(5) z^(3)`, we must have `r_(1) = 2 , r_(2) = 5 and r_(3) = 3.` So, coefficient of `x^(2) y^(5) z^(3) = (10!)/(2!5!3!) 2^(2) xx3^(3)` . |
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| 10. |
The total number of terms in the expansion of `(2x - y + 4z)^(12)`,isA. 90B. 91C. 13D. none of these |
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Answer» Correct Answer - b The total number is `""^(12 + 3-1)C_(3-1) = 91` |
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| 11. |
The numerically greatest term in the expansion of `(1 + x)^(10)` when ` x = 2//3`, isA. `4^(th)`B. `5^(th)`C. `6^(th) `D. `3^(3d)` |
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Answer» Correct Answer - b Let `T_(r) and T_( r+1) ` denote the `r^(th) and (r + 1)^(th)` terms in the expansion of `(1 + x)^(10)`. Then ltbr. `T_(r) = ""^(10)C_(r-1) x^(r-1) and T_(r +1) = ""^(10)C_(r) x^(r)` . `therefore (T_(r +1))/(T_(r)) = (""^(10)C_(r) x^(r))/(""^(10)C_(r-1)x^(r-1))` `rArr (T_(r +1))/(T_(r)) = (""^(10)C_(r) )/(""^(10)C_(r-1))x` `rArr (T_(r +1))/(T_(r)) = (10! )/((10-r)!r!)xx((10 - r+1)!(r-1))/(10!)x` `rArr (T_(r +1))/(T_(r)) = (11-r)/(r)x rArr (T_(r +1))/(T_(r))= ((11 -r)/(r))xx(2)/(3) [because x = 2 //3]` Now ` (T_(r +1))/(T_(r))gt1 rArr ( (11-r)/(r))xx(2)/(3) gt 1 rArr22 gt 5r rArr r lt 4 (2)/(5)` `therefore (4 +1) ^(th) `i.e. 5th term is the greatest term. |
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| 12. |
The greatest term in the expansion of `(1 + 3X)^(54)` when ` x = (1)/(3)`,isA. `28^(th)`B. `25^(th)`C. `26^(th)`D. `24^(th)` |
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Answer» Correct Answer - a Let `T_(r +1) and T_(r)` denote the `(r +1)^(th) and r^(th)` terms respectively. Then `T_(r+1) = ""^(54)C_(r) (3x)^(r) and T_(r) = ""^(54)C_(r-1) (3x)^(r-1)` `therefore (T_(r+1))/(T_(r)) = (""^(54)C _(r))/(""^(54)C _(r-1))xx (3x)^(r) = (54 - r +1)/(r) xx(3x)^(r)` `rArr (T_(r+1))/(T_(r)) = (55-r)/(r) [because x = (1)/(3)]` Now,` (T_(r+1))/(T_(r)) gt 1 rArr (55-r)/(r) rArr r lt 27 (1)/(2)` Hence, `28 ^(th)` term is the greatest term. |
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| 13. |
The sum `sum_(r=0)^(m) ""^(10)C_(r) xx""^(20)C_(m -r)` is maximum ltbr. When m=A. 5B. 10C. 15D. 20 |
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Answer» Correct Answer - c We have, ` sum_(r=0)^(m) ""^(10)C_(r)xx""^(20)C_(m-r)` = `sum_(r=0)^(m)` Coefficient of `x^(r)` in `(1 + x)^(10)` ltbr. `xx` Coefficient of `x^(m-r)` in (1 +x)^(20)` Coefficient of `x^(m)` in `(1 +x)^(30)` `""^(30)C_(m)` is maximum for m = 15. |
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| 14. |
Find the numerically Greatest Term In the expansion of `(3-5x)^15` when x=1/5A. `""^(15)C_(3) xx3^(10)`B. `""^(15)C_(3) xx3^(11)`C. `""^(15)C_(12) xx3^(12)`D. `""^(15)C_(11) xx3^(12)` |
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Answer» Correct Answer - c Let `T_(r +1) and T_(r)` denote the `(r +1)^(th)` term respectively. Then `T_(r-1) = ""^(15)C_(r) 3^(15-r) (-5x)^(r) and , T_(r) = ""^(15)C_(r-1)3^(15 - r +1) (-5x)^(r-1)` `rArr (T_(r+1))/(T_(r)) = (15-r+1)/(r)((-5x)/(3))` `rArr (T_(r+1))/(T_(r)) = (16-r)/(r)xx(-(5)/(3)xx(1)/(5))`, when `x = (1)/(5)` `rArr (T_(r+1))/(T_(r)) = (16-r)/(r)xx(1)/(3) `, numberically [ Neglecting minus sing] Now, `(T_(r) +1)/(T_(r)) xx(1)/(3) gt 1 rArr 16 gt 4r rArr r lt 4` Since 4 is an integer. Therefore , 4th and 5th terms are numberically greatest terms. Now , `T_(4) = T_(3+1) = ""^(15)C_(3) = ""^(15)C_(3) 3^(15-3) (-5x)^(3)` `rArr T_(4) = ""^(15)C_(3) xx3^(12) (-5xx(1)/(5))^(3), when. x = (1)/(5)` `rArr T_(4) = ""^(15)C_(3) xx3^(12)` (numberically) |
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| 15. |
The coefficient of `x^50` in the polynomial `(x + ^50C_0)(x +3.^5C_1) (x +5.^5C_2).....(x + (2n + 1) ^5C_50)`, isA. `50.2^(50)`B. `50.2^(51)`C. `51.2^(50)`D. `50 .(2^(50) +1)` |
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Answer» Correct Answer - c `(x + ""^(50)C_(0)) (x +3. ""^(50)C_(1))(x + 5 . ""^(50)C_(2))...(x + (2b +1)""^(50)C_(50))` = `x^(51) + x^(50) ""^(50){""^(50)C_(0) + 3 . ""^(50)C_(2)+…+ 101""^(50)C_(50)} + …` `therefore `Coefficient of `x^(50)` = `""^(50)C_(0) +3. ""^(50)C_(1) +5. ""^(50)C_(2)+...+101""^(50)C_(50)` =` sum _(r=0)^(50) (2r +1) ""^(50)C_(r)` `= 2 sum_(r=0)^(50) r .""^(50)C_(r) + sum_(r=0)^(50) ""^(50)C_(r)` `= 2xx 50sum_(r=1)^(50) ""^(49)C_(r-1) + sum_(r=0)^(50) ""^(50)C_(r)` `100 xx2^(49) + 2^(50) = 51.2^(50)` . |
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| 16. |
The coefficients of three consecutive terms of `(1+x)^(n+5)`are in the ratio 5:10:14. Then `n=`___________.A. 5B. 7C. 6D. 8 |
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Answer» Correct Answer - c Let `t^(th), (r +1)^(th) and (r +2)^(th)` be three consecutive terms in the expansion of `(1 + x)^(n+5)`. It is given that `""^(n+5)C_(r-1) :""^(n+5)C_(r) :""^(n+5)C_(r+1) = 5:10:14` `rArr (""^(n+5)C_(r))/ (""^(n+5)C_(r-1)) =(10)/(5) and (""^(n+5)C_(r+1))/(""^(n+5)C_(r)) =(14)/(10)` `rArr (n+5 - r +1)/(r) = 2 and ((n+5) -r)/(r+1) = (7)/(5)` `rArr n- 3r + 6 = 0 and 5n - 12r + 18 = 0` `rArr n = 6 , r = 4` |
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| 17. |
The sum of the series `""^(4)C_(0) + ""^(5)C_(1) x + ""^(6)C_(2) x^(2)+""^(7)C_(3)x^(3) +... ` to `infty`, isA. `(1)/((1 - X)^(5)`B. `(1)/( 1 - X)^(5)`C. `(1 + X)^(-5)`D. none of these |
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Answer» Correct Answer - B We know that ` (1 - x)^(-n) = ""^(n-1)C_(0) + ""^(n)C_(0) + ""^(n)C_(1) x + ""^(n +1)C_(2) x^(2) + ..........` Here, we have n= 5. Hence, the sum of the series is `(1- x)^(-5) = (1)/((1 - x)^(5)` |
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| 18. |
Let a, b, c be the sides of `Delta`ABC opposite to angles A, b,C respecitvely. Let `alpha = sum_(r=0)^(n) ""^(n)C_(r) b^(n-r) c^(r) cos{rB - (n-r)C}` and `beta = sum_(r=0)^(n) ""^(n)C_(r) b^(n-r) c^(r) sin{rB - (n-r)C}` Statement -1: `alpha = alpha^(n)` Statement-2: `beta = alpha^(n)`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - c We have, `alpha + i beta = sum_(r=0)^(n) ""^(n)C_(r) b^(n-r) c^(r) e^(i[rB -(n-r)C])` `rArr alpha + i beta= sum_(r=0)^(n) ""^(n)C_(r) (be^(-ic))^(n-r) (ce^(iB))^%(r)` `rArr alpha + i beta= (b e ^(-iC) + ce^(iB))^(n)` `rArr alpha + i beta = {(b cos C + c cos B) + i (-b sin C + csin B)}^(n)` `rArr alpha + i beta = (a + i0)^(n)` "" `[{:(because a = b cos C + c sin B),( " "&(b)/(sinB) = (c)/(sinC)):}]` `rArr alpha + i beta = a^(n)` `rArr alpha = a^(n) and beta = 0.` |
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| 19. |
Let m,`in` N and `C_(r) = ""^(n)C_(r)`, for ` 0 le r len` Statement-1: `(1)/(m!)C_(0) + (n)/((m +1)!) C_(1) + (n(n-1))/((m +2)!) C_(2) +… + (n(n-1)(n-2)….2.1)/((m+n)!) C_(n)` ` = ((m + n + 1 )(m+n +2)…(m +2n))/((m +n)!)` Statement-2: For r `le`0 `""^(m)C_(r)""^(n)C_(0)+""^(m)C_(r-1)""^(n)C_(1) + ""^(m)C_(r-2) ""^(n)C_(2) +...+ ""^(m)C_(0)""^(n)C_(r) = ""^(m+n)C_(r)`.A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - a We have, `""^(m)C_(r)""^(n)C_(0)+""^(m)C_(r-1)""^(n)C_(1) + ""^(m)C_(r-2) ""^(n)C_(2) +...+ ""^(m)C_(0)""^(n)C_(r) = ""^(m+n)C_(r)` = Coefficient of `x^(r)` in `{ (1 + x)^(m) (1 + x)^(n)}` = Coefficient of `x^(r)` in ` (1 + x)^(m + n) = ""^(m+n)C_(r)` So, statement-2 is true. Now, `(1)/(m!) C_(0) + (n)/((m+1)!) C_(1) (n(n-1))/((m+2)!) C_(2) +... + (n(n-1) (n -2)...2.1)/((m +n)!) C_(n)` `= (n!)/((m + n)!) { ((m+n)!)/(m!n!) ""^(n)C_(0) + ((m +n)!)/((m+1)!( n-1)!) ""^(n)C_(1)` ` + ((m + n)!)/((n-2)!) ""^(r)C_(2) +...+ ((m+n)!)/((m+n)!) ""^(n)C_(n)}` `= (n!)/((m+n)!) {""^(m+n)C_(n-2) ""^(n)C_(0) + ""^(m+n)C_(n-1) ""^(n)C_(1) + ""^(m+n)C_(n-2)""^(n)C_(2) +...+""^(m+n)C_(0)""^(n)C_(n)}` `= (n!)/((m+n)!)""^(m+n+n)C_(n)` [Using statement-2] `= (n!)/((m+n)!) xx((m + 2n)!)/((m+n)!n!) ` `=((m +n+1)(m +n+2)(m+ 2n))/((m+n)!) ` So, statement-1 is also true. Statement-2 is a correct expanation for statement-1. |
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| 20. |
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^(21) - 2^(11)`B. `2^(21) - 2^(10)`C. `2^(20) - 2^(9)`D. `2^(20) - 2^(10)` |
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Answer» Correct Answer - d We have `+( ""^(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))` `= ( ""^(21)C_(1)+ ""^(21)C_(2)+ ""^(21)C_(3)+ ""^(21)C_(4) + ...+ ""^(21)C_(10))` `= ( ""^(10)C_(1)+ ""^(10)C_(2)+ ""^(10)C_(3)+ ""^(10)C_(4) + ...+ ""^(10)C_(10))` `=(1)/(2){ ""^(21)C_(1)+2""^(21)C_(2)+ 2""^(21)C_(3)+2""^(21)C_(4)+...+2 ""^(21)C_(10)}` ` -{ ""^(10)C_(1)+ ""^(10)C_(2)+ ""^(10)C_(3)+ ""^(10)C_(4) + ...+ ""^(10)C_(10)}` ` = (1)/(2) {( ""^(21)C_(1)+""^(21)C_(20))+ (""^(21)C_(2)+""^(21)C_(19))+(""^(21)C_(3)+""^(21)C_(18)) ` `+...+(""^(21)C_(10)+""^(21)C_(11))}={""^(10)C_(1)+""^(21)C_(2)` `+""^(10)C_(3)+""^(10)C_(4)+...+ "^(10)C_(10)}` `=(1)/(2) {""^(21)C_(1)+""^(21)C_(2)+""^(21)C_(3)+...+ ""^(21)C_(19) +""^(21)C_(20)}` `-{ ""^(10)C_(1)+ ""^(10)C_(2) + ...+ ""^(10)C_(10)}` `=(1)/(2) {(2^(21)- (""^(21)C_(0)+""^(21)C_(21))}+{2^(10) - ""^(10)C_(0)}` `=(1)/(2) {(2^(21)-2)(2^(10) -1) = 2^(20) - 2^(10)`. |
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| 21. |
If`C_0,C_1,C_2..C_n` denote the coefficients in the binomial expansion of `(1 +x)^n`, then `C_0 + 2.C_1 +3.C_2+. (n+1) C_n`A. `n2^(n)`B. `n2^(n-1)`C. `2^(n-1)`D. `(n-1)2^(n-1)` |
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Answer» Correct Answer - b We have, `C_(1) + 2. C_(2) + 3. C_(3) +...+n C_(n)` `= sum_(r=1)^(n) r. C_(r)` `= sum_(r=1)^(n) r.""^(n) C_(r)" "[because C_(r) = ""^(n) C_(r)] ` `= sum_(r=1)^(n) r.(n)/(r) ""^(n-1)C_(r) [because ""^(n)C_(r) = (n)/(r)""^(n-1)C_(r-1) ]` `= n sum_(r=1)^(n) (n)/(r) ""^(n-1)C_(r-1) ` `=( ""^(n-1)C_(0)+ ""^(n-1)C_(1) + ...+ ""^(n-1)C_(2)+...+""^(n-1)C_(n-1)) = n. 2^(n-1)`. |
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| 22. |
If`C_0,C_1,C_2..C_n` denote the coefficients in the binomial expansion of `(1 +x)^n`, then `C_0 + 2.C_1 +3.C_2+. (n+1) C_n`A. `n2^(n-1)`B. `(n+ 1)2^(n-1)`C. `(n+ 2)2^(n-1)`D. `(n+ 2) 2^(n)` |
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Answer» Correct Answer - c We have, `C_(0) + 2. C_(1) + 3. C_(2) + ... +(n+1) C_(n)` ltbrge `sum _(r=0)^(n) (r + 1) C_(r)` = `sum _(r=0)^(n) (r + 1) ""^(n)C_(r)` `sum _(r=0)^(n) (r .""^(n)C_(r)+ ""^(n)C_(r)) ` `sum _(r=0)^(n) r .""^(n)C_(r)+sum _(r=0)^(n) ""^(n)C_(r)` `sum _(r=0)^(n) r .(n)/(r)""^(n-1)C_(r-1)+sum _(r=0)^(n) ""^(n)C_(r) " "[because ""^(n)C_(r) = (n)/(r). ""^(n-1)C_(r-1) ]` `n(sum_(r=1)^(n) ""^(n-1)C_(r-1))+(sum_(r=1)^(n) ""^(n)C_(r))` ` = n[(""^(n-1)C_(0) + ""^(n-1)C_(1) + ... + ""^(n-1)C_(n-1)]` `+[(""^(n)C_(0) + ""^(n)C_(1) +...+ ""^(n)C_(n))]` `n.2^(n-1)+ 2^(n) = n.2^(n - 1) + 2.2(n-1) = (n +2).2^(n-1)` . |
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| 23. |
If `C_(0), C_(1), C_(2), ..., C_(n)` denote the binomial cefficients in the expansion of `(1 + x )^(n)` , then ` a C_(0) + (a + b) C_(1) + (a + 2b) C_(2) + ...+ (a + nb)C_(n) = `.A. `(a + nb ) 2^(n)`B. `(2a + nb) 2^(n)`C. `(a + nv)2^(n-1)`D. `(2a + nb) 2^(n-1)` |
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Answer» Correct Answer - d We have, ` a C_(0) + (a + b) C_(1) + (a + 2b) C_(2) + ...+ (a + nb)C_(n) ` `sum_(r=0)^(n) (a + rb)""^(n)C_(r)` `sum_(r=0)^(n) a.""^(n)C_(r) + sum_(r=0)^(n) rb""^(n)C_(r)` `=a(sum_(r=0)^(n) ""^(n)C_(r)) +b (sum_(r=0)^(n) r""^(n)C_(r))` `=a(sum_(r=0)^(n) ""^(n)C_(r)) +b (sum_(r=0)^(n) r.(n)/(r)""^(n-1)C_(r-1))` `=a(sum_(r=0)^(n) ""^(n)C_(r)) +b n(sum_(r=0)^(n) ""^(n-1)C_(r-1))` `a. 2^(n) + bn2^(n-1) " " [because sum_(r=0)^(n) ""^(n)C_(r)=2^(n), sum_(r=1)^(n) ""^(n-1)C_(r-1)= 2^(n-1)]` = `(2a + bn) 2^(n-1)` . |
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| 24. |
If `C_(0), C_(1), C_(2), ..., C_(n)` denote the binomial cefficients in the expansion of `(1 + x )^(n)` , then`1^(2).C_(1) + 2^(2) + 3^(3).C_(3) + ...+n^(2).C_(n)=`.A. `(n + 1)2^(n-2)`B. `n(n + 1)2^(n-1)`C. `n(n + 1)2^(n-2)`D. `n(n-1) 2^(n-2)` |
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Answer» Correct Answer - c We have, `1^(2).C_(1) + 2^(2) + 3^(3).C_(3) + ...+n^(2).C_(n)` `= sum_(r=1)^(n) r^(2) .C_(r)` `= sum_(r=1)^(n) r^(2) .""^(n)C_(r)` `= sum_(r=1)^(n) r(r-1) ""^(n)C_(r)+sum_(r=1)^(n) r.""^(n)C_(r)` `= sum_(r=1)^(n) r(r-1) .(n)/(r).(n-1)/(r-1) ""^(n-2)C_(r-2)+sum_(r=1)^(n) r.(n)/(r)""^(n-1)C_(r-1)` ` = n(n -1) ( sum_(r=1)^(n) ""^(n-2)C_(r-2))+n(sum_(r=1)^(n) ""^(n-1)C_(r-1))` `n(n-1)(""^(n-2)C_(0) + ""^(n-2)C_(1) + ...+""^(n-2)C_(n-2)}` `n+{""^(n-1)C_(0) + ""^(n-1)C_(1) + ...+""^(n-1)C_(n-1)}` `n(n-1).2^(n-2) + n^(n-1)` `= n(n-1+2).2^(n-2) = n(n+1)2^(n-2)` |
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| 25. |
If `C_(0),C_(1), C_(2),...,C_(n)` denote the cefficients in the expansion of `(1 + x)^(n)`, then `C_(0) + 3 .C_(1) + 5 . C_(2)+ ...+ (2n + 1) C_(n) = ` .A. `n.2^(n)`B. `(n-1)2^(n)`C. `(n+1)2^(n+1)`D. `( n+1)2^(n)` |
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Answer» Correct Answer - d We have, `c_(0) + 3 C_(1) + 5 C_(2) + ...+(2n +1)C_(n)` `sum_(r=1)^(n) (2r +1) C_(r)` `sum_(r=1)^(n) (2r +1)""^(n)C_(r)" "[because C_(r) = ""^(n)C_(r)`] `sum_(r=1)^(n) (2r. ""^(n)C_(r)+""^(n)C_(r))` `sum_(r=1)^(n) 2r. ""^(n)C_(r)+sum_(r=1)^(n)""^(n)C_(r)` `2sum_(r=1)^(n) r. ""^(n)C_(r)+sum_(r=1)^(n)""^(n)C_(r)` `2sum_(r=1)^(n) r.(n)/(r) ""^(n-1)C_(r-1)+sum_(r=1)^(n)""^(n)C_(r)" "[because ""^(n)C_(r)= (n)/(r) .""^(n-1)C_(r-1)]` `2nsum_(r=1)^(n) ""^(n-1)C_(r-1)+sum_(r=1)^(n)""^(n)C_(r)` `2nsum_(r=1)^(n) 2^(n-1)+2^(n) = n.2^(n) + 2^(n) = (n-1)2^(n)` . |
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| 26. |
The greatest coefficient in the expansion of `(x + y + z + t)^15,` isA. `(15!)/(3!4!)`B. `(15!)/(3!(4!)^(3))`C. `(15!)/(3!(4!)^(2))`D. `(15!)/((3!)^(2)(4!)^(2))` |
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Answer» Correct Answer - b Here,n = 15 and m=4 `therefore q = 3 and r= 3` `therefore ` Greatest coefficient = `(n!)/((q!)^(m-r) [(q+1)!])=(15!)/(3!(4!)^(3))` |
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| 27. |
Find the coefficient of `x^3y^4z^5` in the expansion of `(xy+yz+zx)^6`A. 120B. 20C. 30D. 60 |
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Answer» Correct Answer - d Wehave, `(xy + zyzx)^(6) = sum_(r + s + t )(6!)/(r!s!t!) (xy)^(r)(yz)^(s)(zx)^(t)` `= sum_(r + s + t )(6!)/(r!s!t!) x^(r+t)y^(r+s)z^(s+t)` If the general term in the above expansion contains `x^(3) y^(4) z^(5)`, then `r + t = 3 , r + s = 4 and s + t = 5` Also,` r + s + t = 6` Solving these equations, we get ` r = 1, s = 3, t = 2` ` therefore ` Coefficent of `x^(3) y^(4) z^(5) = (6!)/(1!3!2!) = (6!)/(2!3!) = 60` |
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| 28. |
`1^(2). C_(1) - 2^(2) . C_(2)+ 3^(2). C_(3) -4^(2)C_(4) + ...+ (-1).""^(n-2)n^(2)C_(n)=` |
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Answer» Correct Answer - a We have , `(1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + ... + C_(n) x^(n)` Putting x = -1 on both sides, we get `C_(0) - C_(1) + C_(2) - C_(3) + ...+C_(n) = 0` . |
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| 29. |
If `C_(0), C_(1), C_(2),..., C_(n)` denote the binomial coefficients in the expansion of `(1 + x)^(n)` , then . `1. C_(1) - 2 . C_(2) + 3.C_(3) - 4. C_(4) + ...+ (-1)^(n-1) nC_(n)=` |
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Answer» Correct Answer - a We have , `1. C_(1) - 2 . C_(2)+ 3. C_(3) + ...+ (-1).""^(n-1)nC_(n)` `= sum _(r=1)^(n) (-1)^(r-1) r. ""^(n)C_(r) [becausa C_(r) = ""^(n)C_(n)]` `= sum_(r=1)^(n) (-1)^(r -1)r.(n)/(r) ""^(n-1)C_(r-1)` `= sum_(r=1)^(n) (-1)^(r -1) ""^(n-1)C_(r-1)` `= n xx0 = 0 [ because sum_(r=1)^(r-1)""^(n-1)C_(r-1) = (1 -1)^(n-1) = 0]` . |
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| 30. |
If `C_(0), C_(1), C_(2),..., C_(n)` denote the binomial coefficients in the expansion of `(1 + x)^(n)` , then . `1^(2). C_(1) - 2^(2) . C_(2)+ 3^(2). C_(3) -4^(2)C_(4) + ...+ (-1).""^(n-2)n^(2)C_(n)=`. |
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Answer» Correct Answer - a We have, `1^(2) . C_(1) - 2^(2) . C_(2) + 3^(2) .C_(3) - 4 ^(2) .C_(4) + ...+ (-1)^(n-1) n^(2).C_(n)` `=sum_(r=1)^(n) (-1)^(r-1) r^(2) ""^(n)C_(r) [because ""^(n)C_(r) = C_(r)] ` ` = sum _(r=1)^(n) (-1)^(r-1) [ r (r -1) +r] ""^(2)C_(r)` ` = sum _(r=1)^(n) (-1)^(r-1) r(r-1)""^(n)C_(r) + sum_(r=1)^(n) r (-1)^(r -1)""^(n)C_(r)` `sum _(r=1) ^(n) (-1)^(r -1) r (r -1)(n)/(r). (n-1)/(r-1) ""^(n-2)C_(r - 2) + sum _(r=1)^)(n) r(-1)^(r-1) (n)/(r)n ""^(n-1)C_(r -1)` ` n(n-1)sum_(r=1)^(n) (-1)^(r-1) ""^(n-2) C_(r-2) + n sum_(r=1)^(n) (-1)^(r-1) ""^(n-1)C_(r-1)` `= n(n -1){sum_(r=2)^(n) (-1)^(r -1)""^(n-2)C_(r-2)}+n {sum _(r=1)^(n) (-1)^(r-1) ""^(n-1)C_(r-1)}` `= n(n-1)xx0 + n xx - = 0` . |
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| 31. |
The value of `""(n)C_(1). X(1 - x )^(n-1) + 2 . ""^(n)C_(2) x^(2) (1 - x)^(n-2)` ` + 3. ""^(n)C_(3) x^(3) (1 - x)^(n-3) + ….+ n ""^(n)C_(n) x^(n) , n in ` N isA. nxB. ` n(n -x)`C. ` n (x-1)`D. none of these |
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Answer» Correct Answer - a We have, ` ""^(n)C_(1) . X(1 - x)^(n-1) + 2. ""^(n)C_(2) x^(2) (1 - x)^(n-2)` `+ 3. ""^(n)C_(3) x^(3) (1 -x)^(n-3) + …+ n ""^(n)C_(n) x^(n)` ` = sum_(r=1)^(n) r. ""^(n)C_(r)(1 - x)^(n-r)` ` = sum_(r=1)^(n) r. ""^(n-1)C_(r-1)x^(r) (1 - x)^(n-r)` ` =nx sum_(r=1)^(n) ""^(n-1)C_(r-1)x^(r-1) (1 - x)^((n-1 )-(r-1))` ` nx (x + 1 - x )^(n-1)= nx ` . |
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| 32. |
If `C_(0),C_(1), C_(2),...,C_(N)` denote the binomial coefficients in the expansion of `(1 + x)^(n)` , then `1^(3). C_(1)-2^(3). C_(3) - 4^(3) . C_(4) + ...+ (-1)^(n-1)n^(3) C_(n)=` |
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Answer» Correct Answer - a We have, `1^(3). C_(1)-2^(3). C_(3) - 4^(3) . C_(4) + ...+ (-1)^(n-1)n^(3) C_(n)` `= sum +(r=1)^(n) (-1)^(r-1) r^(3) ""^(n)C_(r)` ` = sum_(r=1)^(n) (-1)^(r-1) {r(r -1) (r-2)+ 3r (r-1) + r} . ""^(n)C_(r)` ` = sum_(r=1)^(n) (-1)^(r-1) r(r -1) (r-2). ""^(n)C_(r)+ 3 sum_(r=1) ^(n) (-1)^(r-1)r(r-1) ""^(n)C_(r) + sum_(r-1)^(n) (-1)^(r-1) r. ""^(n)C_(r)` `sum_(r=1)^(n) (-1)^(r-1)r(r-1) (r-2).(n)/(r).(n-1)/(r-2).(n-2)/(r-2)""^(n-3)C_(r-3)` `+3sum_(r=1)^(n) (-1)^(r-1)r.(n)/(r) ""^(n-1 )C_(r-1)` `= n(n -1) (n-2) {sum_(r=1)^(n) (-1)^(r-3) ""^(n-3)C_(r -3)}` ` - 3 (n-1) sum _(r=1)^(n) (-1)^(r -2)""^(n-2)C_(r-2+ n) {sum_(r=1)^(n) (-1)^(r -1)""^(n-1)C_(r-1)}` `=n(n-1) (n-2)xx0 - 3n(n-1)xx0+n xx- = 0` |
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| 33. |
If `C_(0), C_(1), C_(2),...,C_(n)` denote the binomial coefficients in the expansion of `(1 + x)^n)` , then `xC_(0)-(x -1) C_(1)+(x-2)C_(2)-(x -3)C_(3)+...+(-1)^(n) (x -n) C_(n)=` |
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Answer» Correct Answer - a We have, `xC_(0)-(x -1) C_(1)+(x-2)C_(2)-(x -3)C_(3)+...+(-1)^(n) (x -n) C_(n)=` = `sum_(r=1)^(n) (-1)^(r) (x -r) ""^(n)C_(r)` `x{sum_(x=0)^(n) (-1)^(r) ""^(n)C_(r)} - {sum_(x=0)^(n) (-1)^(r)r ""^(n)C_(r)} ` `x{sum_(x=0)^(n) (-1)^(r) ""^(n)C_(r)} - {sum_(x=0)^(n) (-1)^(r)r.(n)/(r) ""^(n-1)C_(r-1)} ` `x{sum_(x=0)^(n) (-1)^(r) ""^(n)C_(r)} +n {sum_(x=0)^(n) (-1)^(r-1). ""^(n-1)C_(r-1)} ` = `x xx = + n xx 0 = 0` . |
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| 34. |
If `""^(n)C_(0), ""^(n)C_(1),..., ""^(n)C_(n) ` denote the binomial coefficients in the expansion of `(1 + x)^(n) and p + q = 1` , then ` sum_(r=0)^(n) ""r.^(n)C_(r) p^(r) q^(n-r) = `A. nB. npC. npqD. none of these |
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Answer» Correct Answer - b (i) We have, `sum_(r=1)^(n) r. ""^(n)C_(r) p^(r)q^(n-r)` ` sum_(r=1)^(n)r.(n)/(r)""^(n-1)C_(r-1) p.p^(r-1)q^((n-1)-(r -1))` ` = np {sum_(x=1)^(n) ""^(n-1)C_(r-1) p^(r - 1) q^((n-1)-(r-1))}` ` = np (q + p)^(n-1) [ because (q + p)^(n) = sum _(r=0)^(n) ""^(n) C_(r)p^(r) q^(n-r)]` `= np [ because p + q = 1]` . |
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| 35. |
If `""(n)C_(0), ""(n)C_(1), ""(n)C_(2), ...., ""(n)C_(n), ` denote the binomial coefficients in the expansion of `(1 + x)^(n) and p + q =1` ` sum_(r=0)^(n) r^(2 " "^n)C_(r) p^(r) q^(n-r) = ` .A. npqB. np (p+q)C. `np (np + q)`D. ` np (p + nq)` |
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Answer» Correct Answer - b We have, ` sum _(r=0)^(n) r^(2)""^(n)C_(r)p^(r) q^(n-r)` ` sum _(r=0)^(n) [r (r-1)+""^(n)C_(r)p^(r) q^(n-r)` ` sum _(r=0)^(n) r (r-1)""^(n)C_(r)p^(r) q^(n-r) +sum_(r-0)^(n) r. ""^(n)C_(r) p^(r) q^(n-r)` = `n(n - 1){sum_(r=0)^(n) ""^(n-2)C_(r-2) p^(r)q^(n-r) }+n{ sum_(r=0)^(n)""^(n-1)C_(r-1) p^(r)q^(n-r) } ` `= n(n -1) P^(2)` = `n(n - 1){sum_(r=0)^(n) ""^(n-2)C_(r-2) p^(r-2)q^((n-2)-(r-2))}` `+ np {sum_(r=0)^(n) ""^(n-1) C_(r-1) p^(r -1) q^((n-1)-(r-1))}` `= n(n-1) p^(2) (p+q)^(n-2) + np (p + q)^(n -1)` =` n(n -1) p^(2) + np [because p + q = 1]` ` n^(2) p^(2) - np^(2) + np` ` = n^(2) p^(2) + (1 - P ) = n^(2) p^(2) + np q [ because p + q = 1]` |
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| 36. |
If `C_(0), C_(1), c_(2),...,C_(n)` denote the binomial coefficients in the expansion of `(1 + x)^(n)`, then `C_(0) + (C_(1))/(2) + C_(2)/(3) + ...+ (C_(n))/(n+1) or, sum_(r=0)^(n) (C_(r))/(r+ 1)`A. `(2^(n+1) + 1)/(n+ 1)`B. `(2^(n+1) - 1)/(n+ 1)`C. `(2^(n) + 1)/(n+ 1)`D. `(2^(n) - 1)/(n+ 1)` |
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Answer» Correct Answer - b `C_(0) + (C_(1))/(2) + C_(2)/(3) + ...+ (C_(n))/(n+1) ` ` sum_(r=0)^(n) (C_(r))/(r+ 1)` = `sum_(r=0)^(n) (1)/(r+ 1) . ""^(n)C_(r)` `sum_(r=0)^(n) (1)/(r+ 1) .(n+ 1)/(r+ 1). ""^(n)C_(r)` ` (1)/(r+ 1) sum_(r=0)^(n)(n+ 1)/(r+ 1). ""^(n)C_(r)` ` (1)/(r+ 1) sum_(r=0)^(n). ""^(n)C_(r+1) " "[because ""^(n+1)C_(r +1) = (n+ 1)/(r+ 1). ""^(n)C_(r)]` `= (1) (n+1) {""^(n+1)C_(1)+""^(n+1)C_(2)+""^(n+1)C_(3)+...+""^(n+1)C_(n+1)}` `= (1)/(n+1) (""^(n+1)C_(0)+""^(n+1)C_(1)+...+""^(n+1)C_(n+1))-(""^(n+1)C_(0))}` ` (1)/(n+1) (2^(n+1) -1)` |
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| 37. |
If `(1+x)^(n) = C_(0)+C_(1)x + C_(2) x^(2) +...+C_(n)x^(n)` then ` C_(0)""^(2)+C_(1)""^(2) + C_(2)""^(2) +...+C_(n)""^(2)` is equal toA. `2^(2n-2)`B. `2^(n)`C. `((2n)!)/(2(2!)1^(2))`D. `((2n)!)/((n!)^(2))` |
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Answer» Correct Answer - d Using binomial expansion, we have `(1+x)^(n) = C_(0)+C_(1)x + C_(2) x^(2) +...+C_(r)x^(n)+...+C_(n)x^(n)" " ...(i)` `(1+x)^(n) = C_(0)x^(n)+C_(1)x^(n-1) + C_(2) x^(n-2) +...+C_(r)x^(n-r)+...+C_(n-1)x+C_(n)" " ...(ii)` Multiplying (i) and (ii), we get `(1 + x)^(2n) = ( C_(0)+C_(1)x + C_(2)x""^(2) +...+C_(r)x""^(r) + ...+ C_(n)x^(n))` `xx ( C_(0)x^(n)+C_(1)x^(n-1) + C_(2)x""^(n-2) +...+C_(n-r)x""^(r) + ...+ C_(n-1)x+C_(n))` or, `(C_(0) + C_(1) x + C_(2) x^(2) + ...+C_(r)n^(n-2)+...+C_(r) x^(n-r) + ...+C_(n)x^(n))` `xx( C_(0)x^(n)+C_(1)x^(n-1) + C_(2) x^(n-2) +...+C_(r)x^(n-r)+...+C_1)x+C_(n)` `(1 + x)^(2n)`" " ...(iii) Equating the coefficients of `x^(n)` on both sides of (iii) , we get `C_(0)^(2)+C_(1)^(2)+C_(2)^(2)+...+ C_(n)^(2)= ""^(2n)C_(n)` `rArr C_(0)^(2)+C_(1)^(2)+C_(2)^(2)+...+ C_(n)^(2)= ((2n)!)/(n!n!)` . |
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| 38. |
If `C_(0), C_(1), C_(2),..., C_(n)` are binomial coefficients in the expansion of `(1 + x)^(n), ` then the value of `C_(0) - (C_(1))/(2) + (C_(2))/(3) - (C_(3))/(4) +...+ (-1)^(n) (C_(n))/(n+1)` is |
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Answer» Correct Answer - b We have , `C_(0) - (C_(1))/(2) + (C_(2))/(3) - (C_(3))/(4) +...+ (-1)^(n) (C_(n))/(n+1)` `sum_(r=0)^(n) (-1)^(r) (C_(n))/(n+1)` `sum_(r=0)^(n) ((-1)^(r))/(r +1) .""^(n)C_(r)` `sum_(r=0)^(n) ((-1)^(r))/(r +1).(n+1)/(r+1) .""^(n)C_(r)` `= (1)/(n+1) sum_(r=0)^(n) (-1)^(r) .""^(n+1)C_(r+1)" "[because ""^(n+1)C_(r+1)=(n+1)/(r+1).""^(n)C_(r)]` ` (1)/(n+1) sum_(r=0)^(n) ((-1)^(r))/(r +1) .""^(n)C_(r)` `(1)/(n+1)[""^(n+1)C_(1)-""^(n+1)C_(2)+""^(n+1)C_(3)-""^(n+1)C_(4)+...+(-1)^(n) ""^(n+1)C_(n +1)]` `= - (1)/(n+1) [-""^(n+1)C_(1)+""^(n+1)C_(2)-""^(n+1)C_(3)+""^(n+1)C_(4)-...+(-1)^(n) ""^(n+1)C_(n +1)]` `-(1)/(n+1) [""^(n+1)C_(0)-0""^(n+1)C_(1)+""^(n+1)C_(2)-""^(n+1)C_(3)+...+(-1)^(n+1) ""^(n+1)C_(n +1)}-""^(n+1)C_(0)]` `= (1)/(n+1) {0 - ""^(n+1)C_(0)} = (1)/(n+1)` . |
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| 39. |
If `C_(0), C_(1), C_(2),C_(3),..., C_(n)` are the binomial coefficients in the expansion of `(C_(0))/(1)+(C_(2))/(3)+(C_(4))/(5)+(C_(6))/(7)+...,` is equal toA. `(2^(n+1))/(n+1)`B. `(2^(n+1)-1)/(n+1)`C. `(2^(n))/(n+1) `D. none of these |
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Answer» Correct Answer - c From illustration 14 and 15, we have `(C_(0))/(1)+(C_(1))/(2) +(C_(2))/(3) +...+(C_(n))/(n+1) = (2^(n+1)-1)/(n+1)` and `(C_(0))/(1) + (C_(1))/(2) + (C_(2))/(3) - ... +(C_(n))/(n + 1)=(1)/(n+1)` Adding these two, we get `2((C_(0))/(1) + (C_(1))/(2) + (C_(2))/(3) +..... )=(2^(n+1))/(n + 1)` `rArr (C_(0))/(1) + (C_(2))/(3) + (C_(4))/(5) +..... =(2^(n))/(n + 1)`. |
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| 40. |
If `alpha !=1` is an `n^(th)` root of unity and `n in N` such thatfirst three terms in the expansion of `(alpha + x)^n` are `1, alpha and (n - 1)/(2n) bar a^2`, then the value of x, isA. `(1)//(n)`B. `(2)//(n)`C. `1//2`D. `(1)//(4)` |
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Answer» Correct Answer - a Let `alpha = e ^((i2rpi)/(n))`, where 0 `lt r lt n`. Then, `alpha^(n-1) = e ^((i2rpi(n-1))/(n)) = e^((-i2rpi)/(2))= bar(alpha) and alpha^(n-2) = bar(alpha)^(2)` Now, `(alpha + x)^(n) = ""^(n)C_(0) alpha^(n) x^(0) + ""^(n)C_(1) alpha^(n-1) x + ""^(n)C_(2)alpha^b=(n -2) x^(2) +....` `rArr (alpha + x)^(n) = 1 + n bar(alpha) x + (n(n-1))/(2) (bar(alpha))^(2) x^(2) [because alpha^(n) = 1]` `therefore alpha nx = 1 and (n(n-1))/(2) x^(2) = (n-1)/(2n)` `rArr x = (1)/(n)` |
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| 41. |
Find the coefficient of `x^r` in the expansion of `1 +(1 + x) + (1 + x)^2 +.......+ (1 + x)^n`.A. `""^(n)C_(r)`B. `""^(n+1)C_(r )`C. `""^(n+1)C_(r+1)`D. none of these |
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Answer» Correct Answer - c We have, `1 + (1 + x)+(1+x)^(2) +...+ (1 +x)^(n) = (1 - (1 +x)^(n+1))/(1-(1 +x))` `rArr 1 = - (1)/(x) { 1 - (1 - x)^(n+1)}` `rArr 1 = (1)/(x) (1 + x)^(n+1) - (1)/(x)` `therefore ` Coefficient of `x^(r)` in `{ 1 + (1 + x) + (1 + x)^(2) +... + (1 + x)^(n)}` = Coefficient of `x^(r)` in `{( (1 +x)^(n+1))/(x)- (1)/(x)}` ` = ""^(n-1)C_(r+1!)` |
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| 42. |
If `C_(0) , C_(1) , C_(2) ,…, C_(n) ` are coefficients in the binomial expansion of `(1 + x)^(n)` and n is even , then `C_(0)^(2)-C_(1)^(2)+C_(2)^(2)+C_(3)^(2)+...+ (-1)^(n)C_(n)""^(2) ` is equal to . |
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Answer» Correct Answer - b If n is even, suppose `(r + 1)^(th)` term in the binomial expansion of `(1 + x^(2))` contains `x^(n)`. We have, `T_(r +1) = ""^(n)C_(r) (-1)^(r) (x^(2))^(r) = ""^(n)C_(r) (-1)^(r) x^(2r)` For this term to contain `x^(n)` , we must have `2r = n rArr r = n //2` `therefore ` Coefficients of `x^(n) = ""^(n)C_(n//2) (-1)^(n//2)` Also, Coefficient of `x^(n)` in `(1 - x^(2))^(n)` (see illustration 18) is `C_(0)^(2)-C_(1)^(2)+C_(2)^(2)+C_(3)^(2)+...+ (-1)^(n)C_(n)""^(2)` `therefore C_(0)^(2)-C_(1)^(2)+C_(2)^(2)+C_(3)^(2)+...+ (-1)^(n)C_(n)""^(2)= ""^(n)C_(n//2)(-1)^(n//2)` . |
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| 43. |
In the expansion of `(x + a)^(n)` the sum of even terms is E and that of odd terms is O, then OE is equal toA. `(x +a)^(2n) - (x - a)^(2n)`B. `(1)/(4) {(x +a)^(2n) - (x - a)^(2n)}`C. `(1)/(4) {(x +a)^(2n) + (x - a)^(2n)}`D. none of these |
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Answer» Correct Answer - b We have, `(x + a)^(n) = O + E and , (x + a)^(n) = O -E`, ` 40E = (O + E)^(2) - (O - E)^(2)` `rArr 40 E = {( x+a)^(n)}^(2) - {( x - a)^(n) }^(2)` `rArr 40 E = (x +a)^(2n) - (x - a)^(2n)`. |
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| 44. |
If the 4 th term in the expansion of `(a x + (1)/(x))^(n)` is (5)/(2)`, for all x `in` R then the values of a and n are respectivelyA. `(1)/(2),6`B. 1,3C. `(1)/(2),3`D. cannot be found |
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Answer» Correct Answer - a It is given that the fouth term in the expansion of `(ax+ (1)/(x))^(n) is (5)/(2)` `therefore ""^(n)C_(3) (a x)^(n-3) ((1)/(x))(3) = (5)/(2)` `rArr ""^(n)C_(3)a^(n-2) x^(n-6) = (5)/(2)` ….(i) `rArr n-6 = 0 [because ` RHS is independent of x] Putting n=6 in (i) , we get `""^(6)C_(3) a^(3) = (5)/(2) rArr a^(3) = (1)/(8) rArr a = (1)/(2)` Hence , n = 6 a ` = (1)/(2)` . |
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| 45. |
Statement -2: `sum_(r=0)^(n) (-1)^( r) (""^(n)C_(r))/(r+1) = (1)/(n+1)` Statement-2: ` sum_(r=0)^(n) (-1)^(r) (""^(n)C_(r))/(r+1) x^(r) = (1)/((n+1)x) { 1 - (1 - x)^(n+1)}`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - a We have, `sum_(r=0)^(n) (-1)^(r) (""^(n)C_(r))/(r+1)x^(r)` ` = - (1)/(x(x +1)) sum_(r=0)^(n) (-1)^(r+1) (n+1)/(r+1) ""^(n)C_(r) x^(r+1)` `= - (1)/(x (n+1)) {( 1 - x)^(n +1) -1} = (1)/((n+1)x) {1 - (1 - x)^(n +1)}` ` = - (1)/(x(n+1)) {(1 -x)^(n+1) -1} = (1)/((n+1)x) {1 - ( 1 - x)^(n+1)}` So, statement-2 is true Replacing x by 1 in statement-2, we get `sum_(r=0)^(n) (-1)^^(r) (""^(n)C_(r))/(r+1) = (1)/(n+1)` So, statement-1 is also true and stetement-2 is a correct explanation for statement-1. |
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| 46. |
In the expansion of `(x +a)^(n)` the sum of even terms is E and that of odd terms is O, them `O^(2) + E^(2)` is equal toA. `(x +a)^(2n) + (x -a)^(2n)`B. `(1)/(2) {(x +a)^(2n) + (x - a)^(2n)}`C. `(1)/(2) {(x +a)^(2n) - (x - a)^(2n)}`D. `(x + a)^(2n) - (x -a)^(2n)` |
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Answer» Correct Answer - b We have , `(x +a)^(n) = O + E and (x -a)^(n) = =O - E` `therefore O^(2) + E^(2) = (1)/(2) {(O+E)^(2) - (O-E)^(2)} = (1)/(2) {(x+a)^(2n) - (x -a)^(2n)}`. |
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| 47. |
If the third term in the expansion of `(1/x+""_"x"(log)_(10 x))^5`is `1000 ,`then find `xdot`A. 100B. 10C. 1D. `1 sqrt(10)` |
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Answer» Correct Answer - a We have, `T_(3) = 1000` `rArr ""^(5)C_(2) ((1)/(x))^(5-2) (x^(log 10x))^(2) = 1000` `rArr x^(2) log_(10)^(x-3)= 100` `rArr 2 log_(10) x-3 = log_(x) 10^(2)` `rArr 2y - 3 = (2)/(y)` , where y `log_(10) x` `rArr 2y^(2) - 3y - 2 = 0` `rArr (2y+1)(y-2)= 0` `rArr y = 2 or y = 11//2` Now , ` y = 2 rArr log_(10) x = 2 rArr x = 10^(2) = 100` and , `y = - 1//2 rArr log_(10) x = - 1//2 rArr x = 10 ^(-1//2)` But , x `gt` 11, So, x = 100 |
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| 48. |
If the coefficients of `x^-2` and `x^-4` the expansion of `(x^(1/3) +1/(2x^(1/3)))^18`, are `m` and `n` respectively, then `m/n` is equal toA. `(5)/(4)`B. `(4)/(5)`C. 27D. 182 |
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Answer» Correct Answer - d Let `T_(r+1)` be the `(r +1)^(th)` term in the expansion of `(x^(1//3) + (1)/(2x^(1//3)))^(18)`. The `T_(r+1) = ""^(18)C_(r) (x^(1//3))^(18-r) ( (1)/(2x^(1//3)))^(r) = ""^(18)C_(r) x ^(6 (2r)/(3))2^(-r)` ltbr. This will contain `x^(-2)` if, ` 6- (2r)/(3) = -2 rArr r = 12` `therefore m = ` Coefficient of `x^(-2) = ""^(18)C_(12) 2^(-12)` For the coefficient of `x^(-4)`, we must have ` 6 - (2r)/(3) = - 4 rArr r = 15` `because n = ` Coefficient of `x^(-4) = ""^(18)C_(15) 2^(-15)` Hence, `(m)/(n) = (""^(18)C_(12) 2^(-12))/(""^(18)C_(15) 2^(-12))= (18!)/(6!12!)xx(3!15!)/(18!)xx2^(3) = (15xx14xx13xx8)/(6xx5xx4)= 182` |
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| 49. |
Statement-1: `sum_(r=0)^(n) (1)/(r+1) ""^(n)C_(r) = (1)/((n+1)x) {( 1 + x)^(n+1) -1}^(-1)` Statement-2: ` sum_(r=0)^(n) (""^(n)C_(r))/(r+1) = (2^(n+1))/(n+1)`.A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - c We have ` sum_(r=0)^(n) (1)/(r+1) ""^(n)C_(r)x^(r)` ` = - (1)/(x(n+1)) sum_(r=0)^(n) (-1)^(r+1) (n+1)/(r+1) ""^(n)C_(r) x^(r+1)` `=- (1)/(x (n+1)) sum_(r=0)^(n) (-1)^(r+1) ""^(n+1)C_(r+1) x^(r +1)` `= - (1)/(x(n+ 1)) {(1-x)^(x+1) -1} - (1)/((n+1) x) {1-(1 - x)^(n+1)}` So, statement-2 is true. Replacing x by 1 in statement-2, we get `sum_(r=0)^(n) (""^(n)C_(r))/(r+1) = (2^(n+1)-1)/(n+1)` So, statement-2 is false. |
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| 50. |
Find the coefficient of `x^4`in the expansion of `(x//2-3//x^2)^(10)dot`A. `(405)/(256)`B. `(504)/(259)`C. `(450)/(263)`D. none of these |
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Answer» Correct Answer - a Suppose `x^(4)` occours in `(r +1)^(th)` term. We have , `T_(r+1) = ""^(10)C_(r) ((x)/(2))^(10-r) (-(3)/(x^(2)))^(3r) = ""^(10)C_(r) x^(10-3r)(-3)^(r) 2^(r-10)` This will contain `x^(4)`,if `therefore 10 - 3r = 4 rArr 3r = 6 rArr r =2` So, `x^(4)` occours in 3rd and its coefficient is ` ""^(10)C_(r)xx(-3)^(2) xx2^(2-10) =""^(10)C_(2)xx(3^(2))/(2^(8)) = (5xx9xx3^(2))/(2^(8)) = (405)/(256)` . |
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