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Question: In the expansion of the following expression \[1 + \left( {1 + x} \right) + { \left( {1 + x} \righ...

In the expansion of the following
expression 1+(1+x)+(1+x)2+....+(1+x)n1 + \left( {1 + x} \right) + { \left( {1 + x} \right)^2} + .... + { \left( {1 + x} \right)^n} , the coefficient of xk(0⩽k⩽n)  {x_k} \left( {0 \leqslant k \leqslant n} \right) \; is
A. n+1Ck+1{}^{n + 1}{C_{k + 1}}
B.nCK{n_{{C_K}}}
C.nCn−K−1{n_{{C_{n - K - 1}}}}
D.None of these

Explanation

Solution

Hint : To answer this type of problem try to know which type series is given. Find their sum, once sum is found use the formula to find the coefficient of the required x.

Complete step-by-step answer :
Given expression 1+(1+x)+(1+x)2+....+(1+x)n1 + \left( {1 + x} \right) + { \left( {1 + x} \right)^2} + .... + { \left( {1 + x} \right)^n}
The expression is in G. P. and total number of terms are n+1
Suppose the sum of the given series is E
E=1+(1+x)+(1+x)2+....+(1+x)nE = 1 + \left( {1 + x} \right) + { \left( {1 + x} \right)^2} + .... + { \left( {1 + x} \right)^n}
Here the first term of the GP is 1 and the common ratio is 1+x1 + x
Now applying the summation formula of a GP we get,
⇒E=(1+x)n+1−1(1+x)−1\Rightarrow E = \dfrac{{{{ \left( {1 + x} \right)}^{n + 1}} - 1}}{{ \left( {1 + x} \right) - 1}}
Or,
⇒E=x−1(1+x)n+1−1\Rightarrow E = {x^{ - 1}} \\{ { \left( {1 + x} \right)^{n + 1}} - 1 \\}
∴ The coefficient of xk{x_k} in E = The coefficient of xk+1  {x_{k + 1}} \; in x−1(1+x)n+1−1=n+1Ck+1{x^{ - 1}} \\{ { \left( {1 + x} \right)^{n + 1}} - 1 \\} = {}^{n + 1}{C_{k + 1}}
Hence the coefficient of xk{x_k} is n+1Ck+1{}^{n + 1}{C_{k + 1}}
So, the correct answer is “Option A”.

Note : Here in this question it is given that k lies between 0 to n, because n can not be more than n or negative means less than 0. Otherwise this series would not be valid to find the coefficient of xk{x_k} .