Question

If $P_n$ denotes the product of the binomial coefficients in the expansion of ${(1+x)}^n$, then $\frac{P_{n+1}}{P_n}$ equals

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

Answer 1

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

Given: $P_n{=}^n{C}_0{.}^n{C}_1{.}^n{C}_2..{.}^n{C}_n$

Now, $\frac{P_{n+1}}{P_n}=\frac{{}^{n+1}{C}_0{.}^{n+1}{C}_1{.}^{n+1}{C}_2..{.}^{n+1}{C}_{n+1}}{{}^n{C}_0{.}^n{C}_1{.}^n{C}_2..{.}^n{C}_n}$

$\Rightarrow \frac{P_{n+1}}{P_n}{=}^{n+1}{C}_0.\left(\frac{{}^{n+1}{C}_1}{{}^n{C}_0}\right).\left(\frac{{}^{n+1}{C}_2}{{}^n{C}_1}\right)...\left(\frac{{}^{n+1}{C}_n}{{}^n{C}_n}\right){.}^{n+1}{C}_{n+1}$

Since ${}^{n+1}{C}_{r+1}=\frac{n+1}{r+1}{.}^n{C}_r$

$\Rightarrow \frac{P_{n+1}}{P_n}=1\left(\frac{n+1}{1}\right)\left(\frac{n+2}{2}\right)...\left(\frac{n+1}{n}\right)1$

$\therefore \frac{P_{n+1}}{P_n}=\frac{{(n+1)}^n}{n!}$