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Question: The value of the expression \(1 - \frac{n}{1}.\frac{(1 + x)}{1 + nx} + \frac{n(n - 1)}{1.2}\frac{(1...

The value of the expression

1−n1.(1+x)1+nx+n(n−1)1.2(1+2x)(1+nx)2−n(n−1)(n−2)1.2.3(1+3x)(1+nx)3+........1 - \frac{n}{1}.\frac{(1 + x)}{1 + nx} + \frac{n(n - 1)}{1.2}\frac{(1 + 2x)}{(1 + nx)^{2}} - \frac{n(n - 1)(n - 2)}{1.2.3}\frac{(1 + 3x)}{(1 + nx)^{3}} + ........is

A

2

B

1

C

3

D

0

Answer

0

Explanation

Solution

The expression can be divided into two parts as

(1−n11(1+nx)+n(n−1)21(1+nx)2−n(n−1)(n−2)1⋅2⋅3(1+nx)3+⋯ )\left( 1 - \frac{n}{1}\frac{1}{(1 + nx)} + \frac{n(n - 1)}{2}\frac{1}{(1 + nx)^{2}} - \frac{n(n - 1)(n - 2)}{1 \cdot 2 \cdot 3(1 + nx)^{3}} + \cdots \right)+ (−nx1+nx+n(n−1)x(1+nx)2−n(n−1)(n−2)1.2x(1+nx)3+⋯ )\left( \frac{- nx}{1 + nx} + \frac{n(n - 1)x}{(1 + nx)^{2}} - \frac{n(n - 1)(n - 2)}{1.2}\frac{x}{(1 + nx)^{3}} + \cdots \right)

= (1−11+nx)n−nx1+nx(1−(n−1)1(1+nx)+(n−1)(n−2)1.2(1+nx)2+⋯ )\left( 1 - \frac{1}{1 + nx} \right)^{n} - \frac{nx}{1 + nx}\left( 1 - \frac{(n - 1)}{1(1 + nx)} + \frac{(n - 1)(n - 2)}{1.2(1 + nx)^{2}} + \cdots \right)

= (nx1+nx)n−nx1+nx(1−11+nx)n−1\left( \frac{nx}{1 + nx} \right)^{n} - \frac{nx}{1 + nx}\left( 1 - \frac{1}{1 + nx} \right)^{n - 1}

= (nx1+nx)n−(nx1+nx)(nx1+nx)n−1\left( \frac{nx}{1 + nx} \right)^{n} - \left( \frac{nx}{1 + nx} \right)\left( \frac{nx}{1 + nx} \right)^{n - 1}

= 0