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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.
| 101. |
The value of `sum_(r=0)^(40) r""^(40)C_(r)""^(30)C_(r)` isA. `40.^(69)C_(29)`B. `40.^(70)C_(30)`C. `.^(60)C_(29)`D. `.^(70)C_(30)` |
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Answer» Correct Answer - A `underset(r=0)overset(40)sumr.^(40)C_(r).^(30)C_(r)=40underset(r=0)overset(40)sum.^(29)C_(r-1).^(30)C_(r)` `= 40 underset(r=0)overset(40)sum.^(39)C_(r-1).^(30)C_(30-r)` `= 40^(39+30)C_(r+1+30-r)` `= 40.^(69)C_(29)` |
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| 102. |
The value of `sum_(r=1)^(15) (r2^(r))/((r+2)!)` is equal toA. `((17)!-2^(16))/((17)!)`B. `((18)!2^(17))/((18)!)`C. `((16)!-2^(15))/((16)!)`D. `((15)!-2^(14))/((15)!)` |
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Answer» Correct Answer - A `(rxx2^(r))/((r+2)!) = ((r+2-2)2^(r))/((r+2)!)` `= (2^(r))/((r+1)!)-(2^(r+1))/((r+2)!)` `= - ((2^(r+1))/((r+2)!)-(2^(r))/((r+1)!))` `= -(V(r)-V(r-1))` `rArr underset(r=1)overset(15)sum(rxx2)/((r+2)!)=-(V(15)-V(0))` `= - ((2^(16))/(17!)-(2)/(2!))` `= 1-(2^(16))/((17)!)` |
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| 103. |
If `f(x) = .^(40)C_(1).x(1-x)^(39) + 2..^(40)C_(2)x^(2)(1-x)^(38)+3..^(40)C_(3)x^(3)(1-x)^(37)+"….."+40..^(40)C_(40)x^(40)`, then the value of `f(3)` isA. 120B. 150C. 200D. 240 |
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Answer» Correct Answer - A `f(x)=.^(40)C_(1).x(1-x)^(39)+2..^(40)C_(2)xxx^(2)(1-x)^(38)+3..^(40)C_(3)xx x^(3)(1-x)^(37)+"...."+40..^(40)C_(40)xxx^(40)` `T_(r)=r..^(40)C_(r).x^(r)(1-x)^(40-r)` `= 40x..^(39)C_(r-1).x^(r-1)(1-x)^(40-r)` `:. f(x)=40x(x+1-x)^(39)` `= 40x` |
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| 104. |
`(n+2)^n C_0 2^(n+1)-(n+1)^n C_1 2^n^n C_2 2^(n-1)-`is equal to`4`b. `4n`c. `4(n+1)`d. `2(n+2)`A. 4B. 4nC. 4(n+1)D. 2(n+2) |
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Answer» Correct Answer - C `t_(r+1)=(-1)^(r)(n-r+2).^(n)C_(r)2^(n-r+1)` `= (n+2)2^(n+1)(-1)^(r).^(n)C_(r)(1/2)^(r)-2^(n+1)r.^(n)C_(r)(1/2)^(r)` `= (n+2)2^(n+1).^(n)C_(r)(-1/2)^(r) + 2^(n)n.^(n-1)C_(r-1)(-1/2)^(r-1)` `:.` Sum `= (n+2)2^(n+1){.^(n)C_(0) - .^(n)C_(1) xx 1/2 .^(n)C_(2)xx(1/2)^(2)-"...."}` `+2^(n){.^(n-1)C_(0)-.^(n-1)C_(1)xx1/2+.^(n-1)C_(2)xx(1/2)^(2)+"...."}` `= (n+2)2^(n+1)(1-1/2)^(n)+n2^(n)(1-1/2)^(n-1)` `= 2(n+2)+2n` `= 4n+4` |
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| 105. |
The value of `sum_(r=0)^(20)(-1)^(r )(""^(50)C_(r))/(r+2)` is equal toA. `(1)/(50xx51)`B. `(1)/(52xx50)`C. `1/(52xx51)`D. none of these |
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Answer» Correct Answer - C Here, `T_(r)=(-1)^(r)(.^(50)C_(r))/(r+2)` `=(-1)^(r)(r+1)(.^(50)C_(r))/((r+1)(r+2))` `=(-1)^(r)(r+1)(.^(52)C_(r+2))/(51xx52)` `= (-1)^(r)([(r+2)-1]^(52)C_(r+2))/(51xx52)` `= (-1)^(r)([52.^(51)C_(r+1)-.^(52)C_(r+2)])/(51xx52)` `= ([-52.^(51)C_(r+1)(-1)^(r+1)-.^(52)C_(r+2)(-1)^(r+2)])/(51xx52)` `underset(r=0)overset(50)sum(-1)^(r)(.^(50)C_(r))/(r+2)` ` = -52((1-1)^(51)-.^(51)C_(0))/(51xx52)-((1-1)^(52)-.^(52)C_(0)+.^(52)C_(1))/(51xx52)` `= 1/51-1/52` `= 1/(51xx52)` Alternate solution : `(1-x)^(n)=underset(r=0)overset(n)sum.^(n)C_(r)(-1)^(r)x^(r)` or `x(1-x)^(n)=underset(r=0)overset(n)sum(-1)^(r).^(n)C_(r)x^(r+1)` Intergrating both sides withing the limits 0 to 1, we get `underset(0)overset(1)intx(1-x)^(n)dx=underset(r=0)overset(n)sum(-1)^(r)(.^(n)C_(r))/(r+2)` or `underset(r=0)overset(n)sum(-1)^(r)(.^(n)C_(r))/(r+2) = underset(0)overset(1)intx(1-x)^(n)dx` `= underset(0)overset(1)int(1-x)x^(n)dx`(Replace x by `1-x`) `= |(x^(n+1))/(n+1)-(x^(n+2))/(n+2)|_(0)^(1)` `= (1)/(n+1)-(1)/(n+2)` `= (1)/((n+1)(n+2))` Now put `n = 50`. |
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| 106. |
The value of `sum_(r=0)^(20) r(20-r)(""^(20)C_(r))^(2)` is equal toA. `400.^(39)C_(20)`B. `400.^(40)C_(19)`C. `400.^(39)C_(19)`D. `400.^(38)C_(20)` |
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Answer» Correct Answer - D `=underset(r=0)overset(20)sumr(20-r)xx(.^(20)C_(r))^(2)=underset(r=0)overset(20)sumrxx.^(20)C_(r)(20-r)xx.^(20)C_(20-r)` or `underset(r=0)overset(20)sum20.^(19)C_(r-1)xx20xx.^(19)C_(19-r)` `= 400underset(r=0)overset(20)sum.^(19)C_(r-1)xx.^(19)C_(19-r)` `= 400 xx "coefficient of" x^(19) "in" (1+x)^(19)(1+x)^(19)` `= 400 xx .^(38)C_(18)` `= 400 xx .^(38)C_(20)` |
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| 107. |
If `(1+x^(2))^(n) =sum_(r=0)^(n) a_(r)x^(r )= (1+x+x^(2)+x^(3))^(100)`. If `a = sum_(r=0)^(300)a_(r)`, then `sum_(r=0)^(300) ra_(r)` isA. `.^(n)C_(r)`B. `.^(n)C_(r)3^(r)`C. `.^(2n)C_(r )`D. `.^(n)C_(r )2^(r )` |
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Answer» Correct Answer - D `(1-x)^(n)(1+x)^(n)=underset(r=0)overset(n)suma_(r)x^(r)(1-x)^(n)(1-x)^(n-r)` or `(1-x+2x)^(n) = underset(r=0)overset(n)suma_(r)x^(r)(1-x)^(n-r)` or `underset(r=0)overset(n)sum.^(n)C_(r)(1-x)^(n-r)(2x)^(r)=underset(r=0)overset(n)suma_(r)x^(r)(1-x)^(n-r)` Comparing general term, we get `a_(r) = .^(n)C_(r)2^(r)`. |
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| 108. |
The coefficient of `x^(-17)` in the expansion of `(x^4-1/x^3)^(15)` isA. 1365B. -1365C. 455D. -455 |
| Answer» Correct Answer - D | |
| 109. |
The coefficient of `x^(-17)` in the expansion of `(x^4-1/x^3)^(15)` is |
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Answer» Given expansion is `(x^(4) - (1)/(x^(3)))^(15)` Let the term `T_(r + 1)` constains the coefficient of `(1)/(x^(17))` i.e., `x^(-17)` `:. T_(r + 1) = .^(15)C_(r) (x^(4))^(15 - r) (-(1)/(x^(3)))^(r)` `= .^(15)C_(r) x^(60 - 4xr) (-1)^(r) x^(-3r)` `.^(15)C_(r) x^(60 - 7r) (-1)^(r)` For the coefficient `x^(-17)`, `60 - 7r = - 17` `rArr 7 r = 77 rArr r = 11` `rArr T_(11 + 1) = .^(15)C_(11) x^(60 - 77) (-1)^(11)` `:.` Coefficient of `x^(-17) = (-15 xx 14 xx 13 xx 12 xx 11 !)/(11! xx 4 xx 3 xx 2 xx 1)` `= - 15 xx 7 xx 13 - 1365` |
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| 110. |
Find the coefficient of `x^(15)` in the expansion of `(x - x^(2))^(10)` |
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Answer» Given expansion is `(x - x^(2))_(10)` Let the term `T_(r + 1)` is the general term `:. T_(r + 1) = .^(10)C_(r) x^(10 - r) (-x^(2))^(r)` `= (-1)^(r) . .^(10)C_(r). X^(10 - r) . X^(2r)` `= (-1)^(r) .^(10)C_(r) x^(10 + r)` For the coefficient of `x^(15)`, `10 + r = 15 rArr r = 5` `T_(5 + 1) = (-1)^(5) .^(10)C_(5) x^(15)` `:.` Coefficient of `x^(15) = - (10 xx 9 xx 8 xx 7 xx 6 xx 5!)/(5 xx 4 xx 3 xx 2 xx 1 xx 5!)` `= - 3 xx 2 xx 7 xx 6 = -252` |
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| 111. |
The expressions `1+x,1+x+x^2,1+x+x^2+x^3.............1+x+x^2+..............+x^a` are mutiplied together and the terms of the product thus obtained are arranged in increasing powers of `x` in the from of `a_0+a_1x+a_2x^2+.................,` then sum of even coefficients?A. `20!`B. `21!`C. `(21!)/(2)`D. `19!` |
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Answer» Correct Answer - C `(c )` Let `f(x)=(1+x)(1+x+x^(2))(1+x+x^(2)+x^(3))…(1+x+x^(2)+….+x^(20))` Highest degree of `f(x)` `=` highest degree of `x` present in the `f(x)` `=1+2+3+…+20` `=(20(21))/(2)=210` Since in the expansion of `f(x)` degree of `x` from zero to `210`, all present so all terms containing `x^(0)`, `x^(1)`, `x^(2)`.....`x^(210)` will be present `:.` total no. of terms `=210+1=211` `(1+x)(1+x+x^(2))....(1+x+x^(2)+....+x^(20))` `=a_(0)+a_(1)x+a_(2)x^(2)+....+a_(210)x^(210)`......`(i)` Replacing `x` by `1//x`, we get `(1+(1)/(x))(1+(1)/(x)+(1)/(x^(2)))....(1+(1)/(x)+(1)/(x^(2))+....+(1)/(x^(20)))` `=a_(0)+(a_(1))/(x)+(a_(2))/(x^(2))+....+(a_(210))/(x_(210))` Taking `L.C.M` both sides, we get `(x+1)(x^(2)+x+1)....(x^(20)+x^(19)+....+x+1)` `=a_(0)x^(210)+a_(1)x^(209)+...+a_(209)x+a_(210)` Thus `a_(0)+x^(210)+a_(1)x^(209)+....+a_(209)x+a_(210)` Thus `a_(r )=a_(210-r)` `a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+.....` `=(1+x)(1+x+x^(2))...(1+x+x^(2)+...+x^(20))` Putting `x=1` in `(i)` `:. a_(0)+a_(1)+a_(2)+....=21!`......`(ii)` Again putting `x=-1` `a_(0)-a_(1)+a_(2)-a_(3)+a_(4)=0`.........`(iii)` Adding equation `(ii)` and `(iii)`, we have `2[a_(0)+a_(2)+a_(4)+....]=21!` `:.a_(0)+a_(2)+a_(4)+....=(21!)/(2)=` sum of even coefficients |
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| 112. |
The expansion `1+x,1+x+x^(2),1+x+x^(2)+x^(3),….1+x+x^(2)+…+x^(20)` are multipled together and the terms of the product thus obtained are arranged in increasing powers of `x` in the form of `a_(0)+a_(1)x+a_(2)x^(2)+…`, then, Number of terms in the productA. `200`B. `211`C. `231`D. none of these |
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Answer» Correct Answer - B `(b)` Let `f(x)=(1+x)(1+x+x^(2))(1+x+x^(2)+x^(3))…(1+x+x^(2)+….+x^(20))` Highest degree of `f(x)` `=` highest degree of `x` present in the `f(x)` `=1+2+3+…+20` `=(20(21))/(2)=210` Since in the expansion of `f(x)` degree of `x` from zero to `210`, all present so all terms containing `x^(0)`, `x^(1)`, `x^(2)`.....`x^(210)` will be present `:.` total no. of terms `=210+1=211` `(1+x)(1+x+x^(2))....(1+x+x^(2)+....+x^(20))` `=a_(0)+a_(1)x+a_(2)x^(2)+....+a_(210)x^(210)`......`(i)` Replacing `x` by `1//x`, we get `(1+(1)/(x))(1+(1)/(x)+(1)/(x^(2)))....(1+(1)/(x)+(1)/(x^(2))+....+(1)/(x^(20)))` `=a_(0)+(a_(1))/(x)+(a_(2))/(x^(2))+....+(a_(210))/(x_(210))` Taking `L.C.M` both sides, we get `(x+1)(x^(2)+x+1)....(x^(20)+x^(19)+....+x+1)` `=a_(0)x^(210)+a_(1)x^(209)+...+a_(209)x+a_(210)` Thus `a_(0)+x^(210)+a_(1)x^(209)+....+a_(209)x+a_(210)` Thus `a_(r )=a_(210-r)` `a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+.....` `=(1+x)(1+x+x^(2))...(1+x+x^(2)+...+x^(20))` Putting `x=1` in `(i)` `:. a_(0)+a_(1)+a_(2)+....=21!`......`(ii)` Again putting `x=-1` `a_(0)-a_(1)+a_(2)-a_(3)+a_(4)=0`.........`(iii)` Adding equation `(ii)` and `(iii)`, we have `2[a_(0)+a_(2)+a_(4)+....]=21!` `:.a_(0)+a_(2)+a_(4)+....=(21!)/(2)=` sum of even coefficients |
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| 113. |
The expansion `1+x,1+x+x^(2),1+x+x^(2)+x^(3),….1+x+x^(2)+…+x^(20)` are multipled together and the terms of the product thus obtained are arranged in increasing powers of `x` in the form of `a_(0)+a_(1)x+a_(2)x^(2)+…`, then, The value of `(a_(r ))/(a_(n-r))`, where `n` is the degree of the product.A. `2`B. `1`C. `1//2`D. depends on `r` |
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Answer» Correct Answer - B `(b)` Let `f(x)=(1+x)(1+x+x^(2))(1+x+x^(2)+x^(3))…(1+x+x^(2)+….+x^(20))` Highest degree of `f(x)` `=` highest degree of `x` present in the `f(x)` `=1+2+3+…+20` `=(20(21))/(2)=210` Since in the expansion of `f(x)` degree of `x` from zero to `210`, all present so all terms containing `x^(0)`, `x^(1)`, `x^(2)`.....`x^(210)` will be present `:.` total no. of terms `=210+1=211` `(1+x)(1+x+x^(2))....(1+x+x^(2)+....+x^(20))` `=a_(0)+a_(1)x+a_(2)x^(2)+....+a_(210)x^(210)`......`(i)` Replacing `x` by `1//x`, we get `(1+(1)/(x))(1+(1)/(x)+(1)/(x^(2)))....(1+(1)/(x)+(1)/(x^(2))+....+(1)/(x^(20)))` `=a_(0)+(a_(1))/(x)+(a_(2))/(x^(2))+....+(a_(210))/(x_(210))` Taking `L.C.M` both sides, we get `(x+1)(x^(2)+x+1)....(x^(20)+x^(19)+....+x+1)` `=a_(0)x^(210)+a_(1)x^(209)+...+a_(209)x+a_(210)` Thus `a_(0)+x^(210)+a_(1)x^(209)+....+a_(209)x+a_(210)` Thus `a_(r )=a_(210-r)` `a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+.....` `=(1+x)(1+x+x^(2))...(1+x+x^(2)+...+x^(20))` Putting `x=1` in `(i)` `:. a_(0)+a_(1)+a_(2)+....=21!`......`(ii)` Again putting `x=-1` `a_(0)-a_(1)+a_(2)-a_(3)+a_(4)=0`.........`(iii)` Adding equation `(ii)` and `(iii)`, we have `2[a_(0)+a_(2)+a_(4)+....]=21!` `:.a_(0)+a_(2)+a_(4)+....=(21!)/(2)=` sum of even coefficients |
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| 114. |
If `r^(th)`term in the expansion of `(x/3-2/x^2)^(10)`contains `x^4`, then find the value of rA. 2B. 3C. 4D. 5 |
| Answer» Correct Answer - B | |
| 115. |
If `r^[th]` and `(r+1)^[th]` term in the expansion of `(p+q)^n` are equal, then `[(n+1)q]/[r(p+q)]` isA. `(1)/(2)`B. `(1)/(4)`C. `1`D. `0` |
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Answer» Correct Answer - C `(c )` `T_(r )=^(n)C_(r-1)p^(n-r+1)*q^(r-1)` `=T_(r+1)=^(n)C_(r)p^(n-r)*q^(r )` (given) `impliesp^(n-r+1)*q^(r-1)=(n-r+1)/(r )p^(n-r)*q^(r )` `implies(p)/((n-r+1))=(q)/(r )` `implies((n+1)q)/((p+q)r)=1` |
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| 116. |
The number of distinct terms in the expansion of is `(x^(3)+(1)/(x^(3))+1)^(200)` isA. `201`B. `400`C. `401`D. `500` |
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Answer» Correct Answer - C `(c )` `(1+x^(3)+(1)/(x^(3)))^(200)` `=1+^(200)C_(1)(x^(3)+(1)/(x^(3)))+^(200)C_(2)(x^(3)+(1)/(x^(3)))^(2).......^(200)C_(200)(x^(3)+(1)/(x^(3)))^(200)` The `R.H.S` is of the form `k_(0)+k_(1)x^(3)+k_(2)(x^(3))^(2)+....k_(200)(x^(3))^(200)+(l_(1))/(x^(3))+(l_(2))/((x^(3))^(2))+....(l_(200))/((x^(3))^(200))` where `k_(0),k_(1),.....,l_(1),l_(2),....` are all real constants `:.` Number of terms `=1+200+200=401` |
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| 117. |
The middle term in the expansion of `(1 - (1)/(x))^(n) (1 - x)^(n)` isA. `""^(2n)C_(n)`B. `-""^(2n)C_(n)`C. `- ""^(2n)C_(n-1)`D. none of these |
| Answer» Correct Answer - A | |
| 118. |
Find the coefficient of x in the expansion of `[sqrt(1+x^2) - x]^-1` in ascending power of x when `|x| |
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Answer» Correct Answer - D `[sqrt(1+x^(2))-x]^(-1)=(1)/(sqrt(1+x^(2))-x) xx ((sqrt(1+x^(2))+x))/((sqrt(1+x^(2))+x))` `= (sqrt(1+x^(2))+x)/(1+x^(2)-x^(2))=x+sqrt(1+x^(2))=x+(1+x^(2))^(1//2)` `= x+1+1/2 x^(2) + 1/2(-1/2) (x^(4))/(2!) + "….."` Therefore, the coefficient of `x^(4)` is `-1//8`. |
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| 119. |
The value of `sum_(r=0)^(10) (r)""^(20)C_(r)` is equal toA. `20(2^(18)+.^(19)C_(10))`B. `10(2^(18)+.^(18)C_(10))`C. `20(2^(16)+.^(19)C_(11))`D. `10(2^(18)+.^(19)C_(11))` |
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Answer» Correct Answer - A `underset(r=0)overset(10)sum(r).^(20)C_(r) = underset(r=1)overset(10)sum20 xx .^(19)C_(r-1)` `= 20(.^(19)C_(0) + .^(19)C_(1) + "......"+ .^(19)C_(10))` `= 20 (.^(19)C_(0) +.^(19)C_(1) + "......" + .^(19)C_(10))` `= 20(1/2 xx 2^(10) + .^(19)C_(10))` `= 20(2^(18)+.^(19)C_(10))` |
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| 120. |
`sum_(k=1)^(oo) k(1-1/n)^(k-1) =`A. `n(n-1)`B. `n(n+1)`C. `n^(2)`D. `(n+1)^(2)` |
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Answer» Correct Answer - C `underset(k=1)overset(oo)sumk(1-1/n)^(k=1)` `=1-2(1-(1)/(n))^(1)+3(1+1/n)^(2)+"….."` `= 1+2r+3t^(2)+"…"` `= (1-t)^(-2)` `= [1-(1-(1)/(n))]^(-2) = (1/n)^(-2) = n^(2)` |
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| 121. |
Value of `sum_(k=1)^(oo) sum_(r=0)^(k) (1)/(3^(k)) (""^(k)C_(r))` isA. `2/3`B. `4/3`C. `2`D. 1 |
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Answer» Correct Answer - C `underset(k=1)overset(oo)sumunderset(r=0)overset(k)sum(1)/(3^(4))(.^(k)C_(r))=underset(k=1)overset(oo)sum(1/(3^(k))(underset(k=0)overset(k)sum.^(k)C_(r)))` `= underset(k=1)overset(oo)sum((2^(k))/(3^(k)))` `= 2/3+(2/3)^(2)+"....."oo` `= (2//3)/(1-2/3) = 2` |
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| 122. |
The number of distinct terms in the expansion of `(x+y^(2))^(13)+(x^(2)+y)^(14)` isA. `27`B. `29`C. `28`D. `25` |
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Answer» Correct Answer - C `(c )` To get common terms in both the expansions `x^(r_(1))(y^(2))^(13-r_(1))=(y)^(14-r_(1))(x^(2))^(r_(2))` `:.r_(1)=2r_(2)` and `26-2r_(1)=14-r_(2)` `:.r_(1)=8`, `r_(2)=4` `:.` Only one term is common. |
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| 123. |
The sum of `1+n(1-1/x)+(n(n+1))/(2!)(1-1/x)^2+oo`will be`x^n`b. `x^(-n)`c. `(1-1/x)^n`d. none of theseA. `x^(n)`B. `x^(-n)`C. `(1-1/x)^(n)`D. none of these |
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Answer» Correct Answer - A `1+n(1-1/x) + (n(n+1))/(2!) (1-1/x)^(2)+"…."oo` `= 1-n[-(1-1/x)]+(-n(-n-1))/(2!)[-(1-(1)/(x))]^(2)+"....."oo` `= [1-(1-(1)/(x))]^(-n)` `= x^(n)` |
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| 124. |
The total number of terms which are dependent on the value of `x`in the expansion of `(x^2-2+1/(x^2))^n`is equal to`2n+1`b. `2n`c. `n`d. `n+1`A. 2n+1B. 2nC. n+1D. none of these |
| Answer» Correct Answer - B | |
| 125. |
The number of distinct terms in the expansion of`(x+1/x+1/(x^2))^(15)`is/are (with respect to different power of `x`)A. 255B. 61C. 127D. none of these |
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Answer» Correct Answer - B `(x+1/x+x^(2)+1/(x^(2)))^(15)=((x^(3)+x+x^(4)+1)/(x^(2)))^(15)` `= (a_(0)+a_(1)x+a_(2)x^(2)+"....."+a_(60)x^(60))/(x^(50))` Hence, the total number of term is 61. |
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| 126. |
If in the expansion of `(a-2b)^(n)`, the sum of `5^(th)` and `6^(th)` terms is 0, then th e values of `a//b`=A. `(n-4)/(5)`B. `(2(n-4))/(5)`C. `(5)/(n-4)`D. `(5)/(2(n-4))` |
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Answer» Correct Answer - B `T_(5) = .^(n)C_(4)a^(n-4)(-2b)^(4)` and `T_(6) = .^(n)C_(5)a^(n-5)(-2b)^(5)` As `T_(5) + T_(6) = 0`, we get `.^(n)C_(4)2^(4)a^(n-4)b^(4)=.^(n)C_(5)2^(5)a^(n-5)b^(5)` or `(a^(n-4)b^(4))/(a^(n-5)b^(5)) = (n!2^(5))/(5!(n-5)!) .(4!(n-4)!)/(n!2^(4))` or `(a)/(b) = (2(n-4))/(5)` |
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| 127. |
In the expansion of `[(1+x)//(1-x)]^2,`the coefficient of `x^n`will be`4n`b. `4n-3`c. `4n+1`d. none of theseA. `4n`B. `4n-3`C. `4n+1`D. none of these |
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Answer» Correct Answer - A Given term can be written as `(1+x)^(2)(1-x)^(-2)=(1+2x+x^(2))[1+2x+3x^(2)+"....."+(n-1)xxx^(n-2)+nx^(n-1)+(n+1)x^(n)+"...."oo]` Coefficient of `x^(n)` is `(n+1+2n+n-1)=4n`. |
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| 128. |
In the expansion of `(3sqrt4+1/(4sqrt6))^20`A. 3B. 18C. 4D. 16 |
| Answer» Correct Answer - A | |
| 129. |
Using binomial theorem, prove that `6^n-5n`always leaves remainder 1 when divided by 25. |
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Answer» `6^n = (1+5)^n` `= C(n,0)1^n5^0+C(n,1)1^(n-1)5^1+C(n,2)1^(n-2)5^2+C(n,3)1^(n-3)5^3+...C(n,n)1^0 5^n` `6^n=1+5n+25(C(n,2)+C(n,3)5+C(n,4)5^2+...C(n,n)5^(n-2))` `6^n-5n = 1+25k`, where ` k = C(n,2)+C(n,3)5+C(n,4)5^2+...C(n,n)5^(n-2)` As, `k` is always positive, so `25k` will always be divisible by `25`. Thus, if we divide `6^n-5n` by `25`, remainder will be `1`. |
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| 130. |
The coefficient of `x^6` in `{(1+x)^6+(1+x)^7+........+(1+x)^(15)}` isA. `""^(16)C_(9)`B. `""^(16)C_(5)-""^(6)C_(5)`C. `""^(16)C_(6) -1`D. none of these |
| Answer» Correct Answer - A | |
| 131. |
The number of real negative terms in the binomial expansion of `(1+i x)^(4n-2),n in N ,x >0`isA. nB. n+1C. n-1D. 2n |
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Answer» Correct Answer - A `T_(r+1) = .^(4n-2)C_(r)("ix")^(r )` `T_(r+1)` is negative, if `i^(r)` is negative and real. `i^(r ) = - 1` `rArr r = 2, 6, 10,"……"` which form an `A.P.` `0 le r le 4n - 2` `4n - 2 = 2 + (r-1)4` `rArr r = n` The required number of terms is n. |
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| 132. |
Using binomial theorem, prove that `6^n-5n` always leaves remainder 1 when divided by 25. |
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Answer» Using binomial expansion , we get `(6^(n)-5n)=(1+5)^(n)-5n` `={.^(n)C_(0)+.^(n)C_(1) xx5 +.^(n)C_(2)xx(5)^(2)+^(n)C_(3) xx(5)^(3)+^(n)C_(4) xx(5)^(4)+...+.^(n)C_(n) xx5^(n)}-5n` `={1+5n+^nC_(2) xx(5)^(2) +^(n)C_(3) xx(5)^(3)+.^(n)C_(4) xx (5)^(4)+...+.^(n)C_(n) xx5^(n)}-5n` `={1+.^(n)C_(2) xx (5)^(2) + .^(n)C_(3) xx(5)^(3)+.^(n)C_(4) xx(5)^(4)+...+ .^(n)C_(n) xx(5)^(n)}` `=(5)^(2) xx{.^(n)C_(2)+ ^(n)C_(3) xx5 + ^(n)C_(4) xx(5)^(2)+...+^nC_(n) xx5^((n-2)) }+1` ` =25 xx ( " an integar ")` 1. Hence, `(6^(n) - 5n)` when divided by 25 leaves the remainder 1. |
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| 133. |
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. n = 2r is a positive integral mulitple of 3B. n+1C. n-1D. 2n |
| Answer» Correct Answer - A | |
| 134. |
Show that `2^(4n+4)-15n-16`, where `n in N`is divisible by `225.` |
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Answer» `2^(4n+4) - 15n - 16 = 16^(n+1) - 15n - 16` `=(15+1)^(n+1) - 15n - 16` `=C(n+1,0)15^(n+1) + C(n+1,1)15^(n) +... +C(n+1,n-1)15^2+C(n+1,n)15^1+C(n+1,n+1)15^0 - 15n - 16` Now, `C(n+1,0)15^(n+1) + C(n+1,1)15^(n) +... +C(n+1,n-1)15^2` is clearly divisible by `225`. So, we can write it as `225k`. So, our expression becomes, `= 225k +C(n+1,n)15^1+C(n+1,n+1)15^0 - 15n - 16` `=225k+15(n+1)+1-15n-16 ` `=225k+15n+15+1-15n-16` `=225k` `:. 2^(4n+4) - 15n - 16 = 225k ` So, clearly `2^(4n+4) - 15n - 16` is divisible by `225.` |
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| 135. |
Show that `2^(4n+4)-15n-16, ` where n `in` N is divisible by 225. |
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Answer» We have `2^(4n+4)-15n - 16 = 2^(4(n+1))-15n - 16 = 16^((n+1))- 15n -16.` Now, `16^(n+1)-15n-16=(1+15)^(n+1)-15n -16` `=^(n+1) C_(0)+^(n+1) C_(1) xx 15 + ^(n+1)C_(2) xx (15)^(2) + ^(n+1)C_(3) xx(15)^(3)+...+.^(n+1)C_(n+1) xx(15)^(n+1)-15n-16` `= 1+(n+1) xx 15 + .^(n+1)C_(2) xx(15)^(2)+ ^(n+1) C_(3) xx(15)^(3)+...+ (15)^(n+1)-15n-16` `=.^(n+1)C_(2) xx (15)^(2) = .^(n+1)C_(3) xx (15)^(3) +...+(15)^(n+1)` `(15)^(2) xx[.^(n+1)C_(2) + .^(n+1)C_(3) xx 15 +...+(15)^(n-1) ]` `=225 xx` (an integar). Hence, `(2^(4n+4)-15n-16)` is divisible by 225. |
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| 136. |
Prove that `sum_(r=0)^(n) ""^(n)C_(r )sin rx. cos (n-r)x = 2^(n-1) xx sin nx`. |
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Answer» `S = underset(r=0)overset(n)sum.^(n)C_(r)sin rx . Cos (n-r) x` ` = .^(n)C_(0) sin 0x cos n x + .^(n)C_(1) sin x cos (n-1)x` `+ .^(n)C_(2)sin 2x cos (n-2)x"….." + .^(n)C_(n-1)sin (n-1) x cos x` `+ .^(n)C_(n)sin nx cos 0x` Writing the sum in reverse order, we get `S = .^(n)C_(n) sin nx cos 0x + .^(n)C_(n-1) sin(n-1)xcos x` `+ .^(n)C_(n-2)sin(n-2)x cos 2x + "......."` `+ .^(n)C_(1) sinx cos (n-1)x+.^(n)C_(0) sin 0x cos nx` `:. S = .^(n)C_(0) sin nx cos 0x + .^(n)C_(1) sin (n-1)x cos x` `+ .^(n)C_(2) sin(n-2)x cos 2x + "......"` `+ .^(n)C_(n-1) sinx cos(n-1)x + .^(n)C_(n) sin 0x cos x nx` Adding (1) and (2), we get `2S = .^(n)sin(0x+nx) + .^(n)C_(1)sin(x+(n-1)x)` `+ .^(n)C_(2) sin (2x+(n-2)x) + "......" + .^(n)C_(n) sin (nx+0x)` ` = (.^(n)C_(0) + .^(n)C_(1)+.^(n)C_(2)+.^(n)C_(3)+"......"+.^(n)C_(n)) sin nx` `= 2^(n) sin nx` `:. S = 2^(n-1) sin nx` |
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| 137. |
Prove that `2 le (1+ (1)/(n))^(n) lt 3` for all `n in N`. |
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Answer» Let `a_(n) = l(1+(1)/(n))^(n)` For `n = 1, (1+1/n)^(n) = 2` Now, `(1+1/n)^(n) = .^(n)C_(0) + .^(n)C_(1) (1/n) + .^(n)C_(2)(1/n)^(n) + "……." + .^(n)C_(r)(1/n)^(r ) + "……" + .^(n)C_(n)(1/n)^(n)` ` = 1+1+(n(n-1))/(2!) (1)/(n^(2)) + (n(n-1)(n-2))/(3!) = 1/(n^(3)) + "......"` ` + (n(n-1)xx"......"xx2xx1)/(n!) (1)/(n^(n)) " " (1)` ` = 2+(1)/(2!)(1-(1)/(n)) + (1)/(3!)(1+(1)/(n))(1-(2)/(n)) +"....."` `+ (1)/(n!)(1-(1)/(n))(1-(2)/(n))"......"(1-(n-1)/(n))" "(2)` Hence, `a_(n) ge 2 ` for all `n in N`. Also, `a_(n) le 1 +1 + (1)/(2!) + (1)/(3!) + "....." + (1)/(r!) + "......" + (1)/(n!)` Fo` 2 le r le n`, we have `r! = 1 xx 2 xx 3 xx "......" xx r ge 2^(r-1)`. `:. a_(n) le 1 + 1 + 1/2 + 1/(2^(2))+ "......" + (1)/(2^(r-1))+"....."+(1)/(2^(n-1))`. ` = 1+(1-(1//2)^(n))/(1-(1//2))` ` = 1+2(1-1/(2^(n))) = 3 - (1)/(2^(n-1))` `:. a_(n) le 3 - 1/(2^(n-1)) lt 3 AA n ge 1` |
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| 138. |
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)`. |
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Answer» `1/(m!).^(n)C_(0)+(n)/((m+1)!).^(n)C_(1)+(n(n-1))/((m+2)!).^(n)C_(2)+"....."+(n(n-1)xx"...."xx2xx1)/((m+2)!).^(n)C_(n)` `= (n!)/((m+n)!)(((m+n)!)/(m!n!).^(n)C_(0)+((m+n)!n)/((m+1)!n!).^(n)C_(1)+((m+n)!n(n-1))/((m+2)!n!) xx .^(n)C_(2)+"....."+((m+n)!)/(n!)+"....."+((m+n)!)/(n!)(n(n-1)xx"....."xx2xx1)/((m+n)!).^(n)C_(n))` `= (n!)/((m+n)!)(.^(m+n)C_(n).^(n)C_(0)+((m+n)!)/((m+1)!(n-1)!).^(n)C_(1)+((m+n)!)/((m+2)!(n-2)!)xx.^(n)C_(2) + "....."+((m+n)!)/(1)(1)/((m+n)!)(1)/((m+n)!) .^(n)C_(n))`. `= (n!)/((m+n)!)(.^(m+n)C_(n).^(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)!)`[coefficient of `x^(n)` in `(1+x)^(m+n)(1+x)^(n)`] `= (n!)/((m+n)!)`[coefficient of `x^(n)` in `(1+x)^(m+2n)`] `= (n!)/((m+n)!).^(m+2n)C_(n)` `= (n!)/((m+n)!)((m+2n)!)/((m+n)!n!)` `= ((m+2n)!)/((m+n)!(m+n)!)` `= ((m+n+1)(m+n+2)(m+n+3)"...."(m+2n))/((m+n)!)` |
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| 139. |
Find the coefficient of `x` in the expansion of `(1 - 3x + 7x^(2)) (1 - x)^(16)`. |
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Answer» Given, expansion `= (1 - 3x + 7x^(2)) (1 - x)^(16)` `= (1 - 3x + 7x^(2)) (.^(16)C_(0) 1^(16) - .^(16)C_(1) 1^(15) x^(1) + .^(16)C_(2) 1^(14) x^(2) + ....+ .^(16)C_(16) x^(16))` `= (1 - 3x + 7x^(2)) (1 - 16 x + 120 x^(2) + ....)` `:.` Cefficient of `x = - 3 - 16 = - 19` |
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| 140. |
The value of `.^(15)C_(0)^(2)-.^(15)C_(1)^(2)+.^(15)C_(2)^(2)-"...."-.^(15)C_(15)^(2)` isA. 15B. -15C. 0D. 51 |
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Answer» Correct Answer - C As we know that `.^(n)C_(0)-.^(n)C_(1)^(2) + .^(n)C_(3)^(2)+"…."+(-1)^(n).^(n)C_(n)^(2) = 0` . (if n is odd) and in the question `n = 15` (odd). Hence, sum of given series is 0. |
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| 141. |
Find the term independent of `x` in the expansion of `(3x - (2)/(x^(2)))^(15)`. |
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Answer» Given expansion is `(3x - (2)/(x^(2)))^(15)` Let `T_(r + 1)` is the general term. `:. T_(r + 1) = .^(15)C_(r) (3x)^(15 - r) ((-2)/(x^(2)))^(r) = .^(15)C_(r) (3x)^(15 - r) (- 2)^(r) x^(-2r)` `= .^(15)C_(r) 3^(15 - r) x^(15 - 3r) (-2)^(r)` For independent of x, `15 - 3r = 0 rArr r = 5` Since, `T_(5 + 1) = T_(6)` is independent of x. `T_(5 + 1) = .^(15)C_(r) 3^(15 - 5) (-2)^(5)` `= - (15 xx 14 xx 13 xx 12 xx 11 xx 10 !)/(5 xx 4 xx 3 xx 2 xx 1 xx 10 !) . 3^(10).2^(5)` `= -3003.3^(10). 2^(5)` |
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| 142. |
The value of `sum_(r=0)^(10)r""^(10)C_(r)3^(r)(-2)^(10-r)` isA. 20B. 10C. 300D. 30 |
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Answer» Correct Answer - D `underset(r=0)overset(10)sumr^(10)C_(r)3^(r)(-2)^(10-r)= 10 underset(r=0)overset(10)sum.^(9)C_(r-1)3^(r)(-2)^(10-r)` `= 10 xx 3 underset(r=0)overset(10)sum.^(9)C_(r-1)3^(r-1)(-2)^(10-r)` `= 30(3-2)^(9) = 30` |
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| 143. |
Prove that `(""^(2n)C_(0))^2-(""^(2n)C_(1))^2+(""^(2n)C_(2))^2-.....+(-1)^n(""^(2n)C_(2n))^2=(-1)^n.""^(2n)C_(n)` |
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Answer» `(""^(2n)C_(0))^3-(""^(2n)C_(1))^3-(""^(2n)C_(2))^3-.....+(-1)^n""^(2n)C_(2n)x^(2n)....(1)` and `(x+1)^(2n)=""^(2n)C_(0)x^(2n)+""^(2n)C_(1)x^(2n-1)+""^(2n)C_(2)x^(2n-2)+....+""^(2n)C_(2n)...(ii)` Multiplying (i) and (ii) ,we get `(x^2-1)^(2n)=(""^(2n)C_(0)-""^(2n)C_(1)x+....+(-1)^n""^(2n)C_(2n))xx(""^(2n)C_(0)x^(2n)+""^(2n)C_(1)x^(2n-1)+.....+""^(2n)C_(2n))` Now ,cofficient of f`x^(2n)` in RHS `=(""^(2n)C_(0))^2-(""^(2n)C_(1))^2+(""^(2n)C_(2))^2-......+(-1)^n(""^(2n)C_(2n)^2` `therefore "General term in L.H.S "., T_(r+1)=""^(2n)C_(r)(x^2)C_(r)(x^2)^(2n-r)(-1)^r` |
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| 144. |
The number of terms in the expansion of `(2x+3y-4z)^n` isA. n+1B. n+3C. `((n+1)(n+2))/(2) `D. none of these |
| Answer» Correct Answer - c | |
| 145. |
The value of`sum_(r=0)^(10) (-1)^(r).4^(10-r)""^(30)C_(r)""^(30-r)C_(10-r)` is equal toA. `.^(30)C_(10) xx 2^(10)`B. `.^(30)C_(9) xx 4^(10)`C. `.^(30)C_(10) xx 3^(10)`D. `.^(30)C_(9) xx 4^(10)` |
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Answer» Correct Answer - C `underset(r=0)overset(10)sum(-1)^(r).4^(10-r)..^(30)C_(r)..^(30-r)C_(10-r)` ` = underset(r=0)overset(10)sum(-1)^(r)..4^(10-r).(30!)/((30-r)!r!) .((30-r)!)/(20!(10-r)!)` `= (30!)/(20!)underset(r=0)overset(10)sum(-1)^(r).4^(10-r).(1)/(r!(10-r)!)` `= (30!)/(10! xx 20!)underset(r=0)overset(10)sum(-1)^(r).4^(10-r).(10!)/(r!(10-r)!)` `= .^(30)C_(10)underset(r=0)overset(10)sum(-1)^(r).4^(10-r)..^(10)C_(r)` `= .^(30)C_(10)(4-1)^(10)` `= .^(30)C_(10).3^(10)` |
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| 146. |
If `sum_(r=0)^(n) ((r+2)/(r+1))""^(n)C_(r)=(2^(8)-1)/(6)`. then n isA. 8B. 4C. 6D. 5 |
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Answer» Correct Answer - D `underset(r=0)overset(n)sum((r+1)/(r+1)+(1)/(r+1)).^(n)C_(1)` `= underset(r=0)overset(n)sum.^(n)C_(r)+underset(r=0)overset(n)sum(.^(n)C_(r))/(r+1)` `= 2^(n)+(1)/(n+1).underset(r=0)overset(n)sum.^(n+1)C_(r+1)` `= 2^(n)+(2^(n+1)-1)/(n+1)` `= ((n+3)2^(n)-1)/(n+1)` `= (2^(8)-1)/(6)` `rArr n = 5` |
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| 147. |
The number of term in the expansion of `[(2 x + 3y)^(4)]^(7)` is 8 |
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Answer» Given expansion is `[(2x + y^(3))^(4)]^(7) = (2 x + y^(3))^(28)` Since, this expansion has 29 terms So, the given statement is false |
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| 148. |
The term independent of `a`in the expansion of`(1+sqrt(a)+1/(sqrt(a)-1))^(-30)i s``^30 C_(20)`b. `0`c. `^30 C_(10)`d. none of theseA. `.^(30)C_(20)`B. `0`C. `.^(30)C_(10)`D. none of these |
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Answer» Correct Answer - B `(1+sqrt(a) + (1)/(sqrt(a)-1))^(-30)` `= (a/(sqrt(a)-1))^(-30)` `= ((sqrt(a)-1)/(a))^(30)` `= (1)/(a^(30)) (1-sqrt(a))^(30)` `=(1)/(a^(30)){.^(30)C_(0)-.^(30)C_(1)sqrt(a) + "...."+.^(30)C_(30)(sqrt(a))^(30)}` There is no term independent of a. |
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| 149. |
Which term in the expansion of `(2-3x)^(19)` has algebrically the last coefficients ?A. `10^(th)`B. `11^(th)`C. `12^(th)`D. `13^(th)` |
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Answer» Correct Answer - C For algebraically the least coefficient we first find the numerically the greatest term for `x = 1`. Using `r le (n+1)/(1+|(a)/(bx)|)`, we get `rle(19+1)/(1+|2/3|)rArr r le 12` Therefore, `T_(13)` and `T_(12)` have numerically the greatest conefficient. `T_(12)` is algebrically the least as it is followed by negative sign. |
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| 150. |
The value of `sum_(r=1)^(n) (-1)^(r+1)(""^(n)C_(r))/(r+1)` is equal toA. `-1/(n+1)`B. `-(1)/(n)`C. `(1)/(n+1)`D. `(n)/(n+1)` |
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Answer» Correct Answer - D `underset(r=1)overset(n)sum(-1)^(r+1)(.^(n)C_(r))/((r+1))= (1)/(n+1)underset(r=1)overset(n)sum(-1)^(r+1).^(n+1)C_(r+1)` `= 1/(n+1)(0-1+(n+1))= (n)/(n+1)` |
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