Question

The coefficient of $x^k(0\le k\le n)$ in the expansion of
$1+(1+x)+(1+x{)}^2+\cdots +(1+x{)}^n$ is

• A. $(n+1)C_k$
• B. ${}^n{C}_k$
• C. ${}^n{C}_{k+1}$
• D. $(n+1)C_{k+1}$

Answer 1

Answer: (D)

$(n+1)C_{k+1}$

Given expansion is
$1+(1+x)+(1+x{)}^2+\cdots +(1+x{)}^n$

Clearly, it is a GP, with (n+1)terms

$S=\frac{(1+x{)}^{n+1}-1}{x}$

$S=\frac{(1+x{)}^{n+1}}{x}-\frac{1}{x}$

In the above expansion, only the first term will have $x^k$ term.
Now, $(1+x{)}^{n+1}{=}^{n+1}{C}_0{+}^{n+1}{C}_{1}x{+}^{n+1}{C}_2{x}^2+....{+}^{n+1}{C}_k{x}^k{+}^{n+1}{C}_{k+1}{x}^{k+1}+.......$
So, the coefficient of $x^k$ in $\frac{(1+x{)}^{n+1}}{x}-\frac{1}{x}$
is $\left(n+1\right)C_{k+1}$