The middle term in the expansion of (1-(1/x))n (1-xn) in powers of x is

Question

The middle term in the expansion of (1-(1/x))n (1-xn) in powers of x is (A) -2nCn-1 (B) -2nCn (C) 2nCn-1 (D) 2nCn

Tardigrade Admin · Accepted Answer

Given expansion can be re-written as (1-(1/x))n .(1-xn) = (-1)nx-n (1-x)2n Total number of terms will be 2n + 1 which is odd (∵ 2n is always even) ∴ Middle term = (2n+1+1/2) = (n+1) th Now, Tr+1= nCr(1)r xn-r So, (2nCn . x2n-n/xn.(-1)n) = 2nCn .(-1)n Middle term is an odd term. So, n + 1 will be odd. So, n will be even. ∴ Required answer is 2nCn.

Q. The middle term in the expansion of $(1 - \frac{1}{x})^{n} (1 - x^{n})$ in powers of x is

$-^{2 n} C_{n - 1}$

$-^{2 n} C_{n}$

$^{2 n} C_{n - 1}$

$^{2 n} C_{n}$

Q. The middle term in the expansion of (1−x1​)n(1−xn) in powers of x is

−2nCn−1​

−2nCn​

2nCn−1​

2nCn​

Q. The middle term in the expansion of $(1 - \frac{1}{x})^{n} (1 - x^{n})$ in powers of x is

$-^{2 n} C_{n - 1}$

$-^{2 n} C_{n}$

$^{2 n} C_{n - 1}$

$^{2 n} C_{n}$