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101.

If the sum of the coefficients in the expansion of `(b + c)^(20) {1 +(a -2) x}^(20)` is equal to square of the sum of the coefficients in the expansion of `[2 bcx - (b + c)y]^(10)`, where a, b, c are positive constants, thenA. ` ge sqrt((a c)`B. `(b +c)/(2) ge a`C. c, a and b are in G. PD. `(1)/(c),(1)/(a),(1)/(b)` are in H.P

Answer» Correct Answer - b
The sum of the coefficients in the expansion of
`(b + c )^(20){1 + (a - 2)x}^(20)` is given by
`S_(1) = ( b + c)^(20) (a - 1)^(20)` [ Putting x = 1]
The sum of the coefficients in the expansion of
`[2 bcx - (b + c)y]^(10)` is given by
`S_(2) = [ 2 bc - (b + c)]^(10)` [ Putting x = y = 1]`
If is given that
`S_(1) = S_(2)""^(2)`
`rArr (b + c)^(20) (a -1)^(20) = (2bc - b - c )^(20)`
`rArr (b + c ) (a -1) - =2bc - b - c`
`rArr a - 1 = (2bc)/(b +c) - 1`
`rArr a= (2 bc)/(b+c)`
`rArr a ` is the H.M of of b and c.
But, `(b +c)/(2)` is the A. M. of b and c. Therefore,
A. M. `ge` H.M.`rArr (b+c)/(2) ge a` .
Discussion
102.

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.A. 4B. 3C. 12D. 6

Answer» Correct Answer - a
We have,
s`""^(n)C_(r) (root(13)(a))^(a-1) ((a)/(sqrt(a^(-1))))^(1) = 14 a^(5//2)`
`rArr na^((n+1)/(13)+(3)/(2)) = 14 a^(5//2) rArr n=14`
`therefore (""^(n)C_(3))/(""^(n)C_(2))=(n-2)/(3) = (14-2)/(3) = 4 [because (""^(n)C_(3))/(""^(n)C_(2))=(n-r+1)/(r)]` s
Discussion
103.

The coefficients of `x^(13)` in the expansion of ` (1 - x)^(5) (1 + x + x^(2) + x^(3) )^(4)` , isA. 4B. -4C. 9D. none of these

Answer» Correct Answer - a
We have,
`(1- x)^(5) (1 + x + x^(2) + x^(3) )^(4)`
`= (1 x)^(5) (1 | x)^(4) (1 | x^(2))^(4)`
` = (1 - x) (1 - x^(2))^(4) (1 + x^(2))^(4)`
` = (1 -x )(1 - x^(4))^(4)`
` = (1 - x) (""^(4)C_(0) - ""^(4)C_(1)x^(4) + ""^(4)C_(2) x^(8) - ""^(4)C_(3)x^(12) + ""^(4)C_(4) x^(16))`
`therefore ` Coefficient of ` x^(13) = ""^(4)C_(3) = 4` .
Discussion
104.

Prove that in the expansion of `(1+x)^n(1+y)^n(1+z)^n`, the sum of the coefficients of the terms of degree `ri s^(3n)C_r`.A. `(""^(n)C_(r))^(3)`B. `3 . ""^(n)C_(r)`C. `""^(3n)C_(r)`D. `""^(n)C_(3r)`

Answer» Correct Answer - c
The given expansion can be written as
`ubrace ({( 1+ x) (1 + x) (1 + x)...(1 +x)})_("n - factors") ubrace({( 1 +y)(1 +y) (1 +y)...(1 +y)})_("n - factors")`
`ubrace({( 1 +z)(1 +z) (1 +z)...(1 +z)})_("n - factors")`
There are 3n factors in this product .
To get a term of degree r. we choose r factors out of these 3n
factors and then multiply second terms in each factor. there
are `""^(3n)C_(r)` such terms each having coefficient 1.
Hence, the sum of the coefficients of `""^(3n)C_(r)`.
Discussion
105.

In the binomial expansion of `(1+a)^(m+n)`, prove that the coefficient of `a^m a n d a^n`are equal.A. A= BB. mA = nBC. nA = mBD. A = 2B

Answer» Correct Answer - a
In the binomial expansion of `(1 +a)^(m+n)`, we have
A = Coefficient of `a^(m) = ""^(m+n)C_(m) = ((m+n)!)/(m!n!)`
and
B = Coefficient of `a^(n) = ""^(m+n)C_(n) = ((m +n)!)/(m!n!)`
Clearly, A = B
Discussion
106.

The value of `{(sqrt(2) +1)^(5) + (sqrt(2) -1)^(5)}` ,isA. 58B. `58sqrt(2)`C. 42D. `42sqrt(2)`

Answer» Correct Answer - b
Using binomial expansion, we have
`(sqrt(2) +1)^(5) + (sqrt(2)-1)^(5) = 2 {(sqrt(2))^(5) +10(sqrt(2))^(3) + 5sqrt(2)} `
`(sqrt(2) +1)^(5) + (sqrt(2)-1)^(5) = 2{4sqrt(2) +20sqrt(2) + 5sqrt(2)} = 5sqrt(2)`.
Discussion
107.

Show that the middle term in the expansion of `(1+x)^(2n)i s((1. 3. 5 (2n-1)))/(n !)2^n x^n ,w h e r en`is a positive integer.A. `(1* 3*5…(2n -1))/(n!) 2^(n) . X^(n)`B. `(1 * 3 * 5 …(2n -1))/(n!!) `C. `""^(2n)C(n)`D. `""^(n)C_(n-1) x^(n-1)`

Answer» Correct Answer - a
We given expansion is `(1 +x)^(2n)` . Here , the index
2n is even
So, `((2n)/(2) +1)^(th)` i.e. `(n+1)^(th)` term is the midle term
`therefore ` Middle = `T_(n+1)`
=` ""^(2n)C_(n) (1)^(2n-n) x^(n) = ""^(2n)C_(n) x^(n) = ((2n)!)/((2n -n)!n!) x^(n)`
`=(1*2*3*4*5*6...(2n -3)(2n -2)(2n-2)(2n-1) (2n))/(n!n!)x^(n)`
`=({1*3*5...(2n -3)(2n -1)}{*2*4*6...(2n-2)(2n)})/(n!n!)x^(n)`
`=({1*3*5...(2n -3)(2n -1)}n!.2^(n))/(n!n!)x^(n)`
`=(1*3*5...(2n -1))/(n!n!)2^(n)x^(n)`.
Discussion
108.

The greatest integer less than or equal to `(sqrt2+1)^6` isA. 197B. 198C. 196D. 199

Answer» Correct Answer - a
Let `(2sqrt(2) +1)^(6) = I + F` where I `in N and 0 lt F lt 1`.
Clearly , I is the greatest integer less then or equal to `(2 sqrt(2) +1)^(6)`
Let G =` (sqrt(2) -1)^(6)` . Then ,
`I + F + G = (sqrt(2) + 1)^(6) + (sqrt(2) -1)^(6)`
`rArr I + F + G = 2 {""^(6)C_(0) (sqrt(2))^(6) + ""^(6)C_(2) (sqrt(2))^(6-2) +....}`
`rArr I + F + G ` = An even integer
`rArr F + G = 1`
Again, `I + F + G`
`2={""^(6)C_(0)(sqrt(2))^(6) + ""^(6)C_(2) (sqrt(2))^(6-2) + ""^(6)C_(4) (sqrt(2))^(6-4) + ""^(6)C_(6) (sqrt(2))^(6-6)}`
`rArr I + 1 = 2 (8 + 15 xx4 + 15 xx4 + 15 xx2 +1) [because F + G = 1 ]`
`rArr I = 197`
Discussion
109.

If A and B are the coefficients of `x^n` in the expansion `(1 + x)^(2n)` and `(1 + x)^(2n-1)` respectively, thenA. A= BB. 2A = BC. A = 2BD. none of these

Answer» Correct Answer - c
We have,
A = Coefficient of `x^(n)` in `( 1 + x)^(2n)`
`rArr A = ""^(2n)C_(n)`
`rArr A = (2n)/(n) ""^(2n-1)C_(n-1) = 2 ""^(2n-1)C_(n-1) [because ""^(n)C_(r) = (n)/(r) ""^(n-1)C_(n-1) ]` and
B = Coefficient of `x^(n) ` in `(1 + x)^(2n-1)`
`rArr B= ""^(2n -1)C_(n) = ""^(2n-1) C_(n-1) [because ""^(n)C_(r) = ""^(n)C_(n-r)]`
Clearly, A = 2B.
Discussion
110.

For a positive integer n if the mean of the binomial coefficients in the expansion of `(a + b)^(2n - 3)` is 16 Then n is equal toA. 4B. 5C. 7D. 9

Answer» Correct Answer - b
The binomail ceofficients in the expansion of
`(a + b)^(2n-3) are ""^(2n-3)C_(0)""^(2n-3)C_(1),...,... ""^(2n-3)C_(2n-3)`.
It is given that their mean is 16
`therefore (""^(2n-3)C_(0)+""^(2n-3)C_(1) + .........+ ""^(2n-3)C_(2n-3))/(2n-2)= 16`
`rArr 2^(2n-3) = 16 (2n -2)`
`rArr 2^(2n-8) = 11 -1`
`rArr n= 5` (By inspection)
Discussion
111.

Show that the coefficient of the middle term in the expansion of `(1+x)^(2n)`is equal to the sum of the coefficients of two middle terms in the expansion of `(1+x)^(2n-1)`.A. A+B=CB. B+C=AC. C+A= BD. A+B +C = 0

Answer» Correct Answer - b
Clearly, `(n+1)^(th)` term is the middle term in the
expansion of `(1 + x)^(2n)` and its coefficients is `""^(2n)C_(n)` i.e. A = `""^(2n)C_(n)`
In the expansion of `(1 +x)^(2n-1), n^(th) and (n +1)^(th)` terms are
middle terms and their coefficient are `""^(2n)C_(n-1) and ""^(2n-1)C_(n)`
respectively i.e. B `""^(2n-1)C_(n-1) and C = ""^(2n)C_(n)`,
We have,
`""^(2n-1)C_(n-1)+""^(2n-1)C_(n) = ""^((2n-1)+1)C_(n)[because ""^(n)C_(r-1) +""^(n)C_(r) = ""^(n+1)C_(r)]`
`therefore B + C = A`
Discussion
112.

The sum of the series `""^(3)C_(0)- ""^(4)C_(1) . (1)/(2) + ""^(5)C_(2)((1)/(2))^(2) - ""^(6)C_(3)((1)/(2))^(3) + ...` to `infty`, isA. 16B. 8C. `(16)/(81)`D. none of these

Answer» Correct Answer - c
We have,
`(1 + x)^(-n) = ""^(n-1)C_(0) - ""^(n)C_(1) x + ""^(n-1)C_(2) x + ""^(n+1)C_(2) x^(2) - ""^(n+2)C_(3) x^(3) + ...`
On comparing the two series, we get
n-1 = 3 a d` x = (1)/(2) rArr n=4 and x = (1)/(2)`
Hence, sum of the given series = `(1 + (1)/(2))^(-4)` = (16)/(81)` .
Discussion
113.

The sum of the series `1 + (1)/(1!) ((1)/(4)) + (1.3)/(2!) ((1)/(4))^(2) + (1.3.5)/(3!) ((1)/(4))^(3)+ ... ` to `infty`, isA. `sqrt(2)`B. 2C. `(1)/(sqrt(2))`D. none of these

Answer» Correct Answer - a
Comparing the given series with the series
`(1 + x)^(n) = 1 + (x)/(1!) + (n(n-1))/(2!)x^(2) + …., ` we get
`nx = (1)/(4) and (n(n-1))/(2!) x^(2) = (1.3)/(2!)((1)/(4))^(2)`
`rArr nx = (1)/(4) and n^(2) x^(2) - (nx) x = (3)/(16)`
`rArr (1)/(16) - (x)/(4) = (3)/(16) rArr (x)/(4) = - (1)/(8) rArr x = - (1)/(2)`
` therefore nx = (1)/(4) rArr n = - (1)/(2)`
Hence, sum of the series = `(1 - (1)/(2))^(-1//2) = sqrt(2)`
Discussion
114.

The approximate value of `(1.0002)^3000`, isA. 1.6B. 1.4C. 1.8D. 1.2

Answer» Correct Answer - a
We have ,
` (1 + x)^(n) cong1 + nx if x lt 1`
`therefore (1 .0002)^(3000) cong1 + 3000 xx0.0002 = 1.6`
Discussion