Question

If $P_n$ denotes the product of all the coefficients in the expansion of $(1+x{)}^n$, then $\frac{P_{n+1}}{P_n}$ is equal to:

• A. $\frac{(n+2{)}^n}{n!}$
• B. $\frac{(n+1{)}^{n+1}}{n+1!}$
• C. $\frac{(n+1{)}^{n+1}}{n!}$
• D. $\frac{(n+1{)}^n}{n+1!}$

Answer 1

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

$(1+x{)}^n$

$P_{n+1}{=}^{n+1}{C}_0{\times }^{n+1}{C}_1{\times }^{n+1}{C}_2\times ........{\times }^{n+1}{C}_{n+1}{P}_n{=}^n{C}_0{\times }^n{C}_1{\times }^n{C}_2\times ........{\times }^n{C}_n$

${\prod }_{r=0}^n\frac{{}^{n+1}{C}_r}{{}^n{C}_r}\left({\because }^{n+1}{C}_{n+1}=1\right)={\prod }_{r=0}^n\frac{(n+1)!}{r!(n+1-r)!}\times \frac{r!(n-r)!}{n!}={\prod }_{r=0}^n\frac{(n+1)}{(n-r+1)}={(n+1)}^n{\prod }_{r=0}^n\frac{1}{n-r+1}=\frac{{(n+1)}^n}{(n+1)!}$

Hence, the correct option is $D$.