Question

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

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

Answer 1

Answer: (C)

$\frac{(n+1{)}^n}{n!}$

$p_n={}^n{C}_0.{}^n{C}_1.{}^n{C}_2......{.}^n{C}_n$

$P_{n+1}={}^{n+1}{C}_0.{}^{n+1}{C}_1.{}^{n+1}{C}_2....{.}^{n+1}{C}_{n+1}$

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

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

$=\left(\frac{\frac{n+1}{1}{.}^n{C}_0}{{}^n{C}_0}\right).\left(\frac{\frac{n+1}{2}{.}^n{C}_1}{{}^n{C}_1}\right)......\left(\frac{\frac{n+1}{n+1}{.}^n{C}_n}{{}^n{C}_n}\right)$

$=\frac{n+1}{1}.\frac{n+1}{2}.....\frac{n+1}{n+1}$

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