Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+...+C_n{x}^n,$then $\frac{(C_0+C_1)(C_1+C_2)...(C_{n-1}+C_n)}{C_1{C}_2..C_n}$ equals-

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

Answer 1

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

$\begin{array}{c}\frac{\left(C_0+4\right)\left(4+C_2\right)\left(C_n-1+C_n\right)}{C_1,C_2{C}_n}\\ \left(1+\frac{1}{\frac{4}{C_0}}\right)\left(1+\frac{1}{\frac{C_2}{4}}\right)\cdots \cdots \cdots \cdots \cdots \cdots \left(1+\frac{1}{\frac{C_n}{C_n-1}}\right)\\ \frac{C_r}{C_r-1}=\frac{n-r-1}{r}\\ \frac{C_1}{C_2}=\frac{n}{1},\frac{C_2}{C_1}=\frac{n-1}{2},\frac{C_3}{C_2}=\frac{n-2}{3}\\ \left(1+\frac{1}{n}\right)\left(1+\frac{1}{\frac{n-1}{2}}\right)\left(1+\frac{1}{\frac{n-2}{3}}\right)...............\left(1+\frac{n}{1}\right)\\ \frac{(n+1)}{n}\frac{\left(n+1\right)}{n-1}\frac{\left(n+1\right)}{(n-2)}...............\left(\frac{n+1}{1}\right)\\ =\frac{(n+1{)}^n}{n!}\end{array}$

Hence, this is the answer.