Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\cdots +C_n{x}^n$, and $(C_0+C_1)(C_1+C_2)\cdots (C_{n-1}+C_n)=k\cdot C_1{C}_2{C}_3\cdots C_n$
then $k$ is equal to

• A. $\frac{(n+1{)}^n}{n!}$
• B. $\frac{n^n}{(n-1)!}$
• C. $\frac{(n+1{)}^n}{(n-1)!}$
• D. None of these

Answer 1

The correct option is A $\frac{(n+1{)}^n}{n!}$

We have,
$C_0+C_1={}^{n+1}{C}_1,C_1+C_2={}^{n+1}{C}_2,\dots ,$
$C_{n-1}+C_n={}^{n+1}{C}_n$
$\therefore \text{The given expression}={}^{n+1}{C}_1{}^{n+1}{C}_2\dots {}^{n+1}{C}_n\cdots \left(1\right)$
$\text{Now,}{}^{n+1}{C}_r=\frac{(n+1)!}{r!(n-r+1)!}=\frac{(n+1)n!}{r!(n-r+1)(n-r)!}\Rightarrow {}^{n+1}{C}_r=\frac{n+1}{n-r+1}{}^n{C}_r\cdots \left(2\right)$

Putting $n=1,2,3,\dots ,n$ in Eq. $\left(2\right)$, we get
$(C_0+C_1)(C_1+C_2)\dots (C_{n-1}+C_n)=\frac{n+1}{n}{C}_1\cdot \frac{n+1}{n-1}{C}_2\dots \frac{n+1}{1}{C}_n=\frac{(n+1{)}^n}{n!}{C}_1{C}_2{C}_3\dots C_n$