$ \frac{1}{1!(n-1)!} – \frac{1}{3!(n-3)!} + \frac{1}{5!(n-5)!} – …is \; equal \; to $

Q: $\large \frac{1}{1!(n-1)!} – \frac{1}{3!(n-3)!} + \frac{1}{5!(n-5)!} – …is \; equal \; to $

(A) $\large \frac{2^{n-1}}{n!}$ for all n ∈ N

(B) $\large \frac{2^{n-1}}{n!}$ for odd values of n only

(C) $\large \frac{2^{n-1}}{n!}$ for even values of n only

(D) none of these

Sol. Given expression $\large = \frac{1}{n!}(n_{C_1} – n_{C_3} + n_{C_5} -…..) = \frac{2^{n/2}sin\frac{n\pi}{4}}{n!}$

Given the integers r > 1, n > 2 , and coefficients of (3r)th and (r+2)nd term in the binomial expansion of (1+x)2n are equal, then

Q: Given the integers r > 1, n > 2 , and coefficients of (3r)th and (r+2)nd term in the binomial expansion of (1+x)2n are equal, then

(A) n = 2r

(B) n =3r

(C) n = 2r+1

(D) None of these

Sol. Coefficients of (3r)th and (r+2)nd terms will be 2nC3r-1 and 2nCr+1

These are equal ⇒ (3r – 1) + (r + 1) = 2n ⇒ n = 2r

Hence (A) is the correct answer.

The value of (n + 2) C0 2n+1 – (n + 1) C12n + n. C2 2n-1 – … is equal to

Q: The value of (n + 2) C0 2n+1 – (n + 1) C12n + n. C2 2n-1 – … is equal to

(A) 4 (1 + n)

(B) 4n

(C) 2n

(D) 2n + 4

Sol. (x – 1)n = C0 xn – C1 xn-1 + Cnxn-2 – ….

x2 (x – 1)n = C0 xn+2 – C1 xn+1 + C2 xn -….

differentiating with respect to x , we get

2x (x – 1)n + x2n (x – 1)n-1 = (n + 2) C0xn+1 – (n + 1) C1 xn + nC2xn-1 – ………

put, x = 2, we get

(n + 2) C0 xn+1 – (n + 1) C12n + n. C22n-1 – …. = 4 + 4n

Hence (A) is the correct answer.

The term independent of x in $ ( \sqrt{\frac{x}{3}} + \sqrt{\frac{3}{2x^2}} )^{10} $ is

Q: The term independent of x in $\large (\sqrt{\frac{x}{3}} + \sqrt{\frac{3}{2x^2}} )^{10} $ is

(A) 1

(B) 5/12

(C) 10C1

(D) None of these

Sol. General term in the expansion is

$\large = 10_{C_r} (\frac{x}{3})^{r/2} (\frac{3}{2x^2})^{\frac{10-r}{2}}$

$\large = 10_{C_r} x^{\frac{3r}{2}-10} . \frac{3^{5-r}}{2^{(10-r)/2}}$

For constant term, 3r/2 = 10  ⇒ r = 20/3

which is not an integer. Therefore, there will be no constant term.

Hence (D) is the correct answer.