Question

If $(1+x{)}^n=C_o+C_{1}x+C_2{x}^2+...+C_n{x}^n$. Then, $C_o{C}_1+C_1{C}_2+...+C_{n-1}{C}_n$ is equal to

• A. $\frac{2n!}{(n-1)!(n+1)!}$
• B. $\frac{(2n-1)!}{(n-1)!(n+1)!}$
• C. $\frac{(2n)!}{(n+2)!(n+1)!}$
• D. None of these

Answer 1

The correct option is A

$\frac{2n!}{(n-1)!(n+1)!}$

To find ${\mathit{C}}_{\mathbf{o}}{\mathit{C}}_{\mathbf{1}}\mathbf{+}{\mathit{C}}_{\mathbf{1}}{\mathit{C}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{C}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{C}}_{\mathbf{n}}$:

Step 1: Find $(x+1{)}^n$

Given that $(1+x{)}^n=C_o+C_{1}x+C_2{x}^2+...+C_n{x}^n$. Then we have to find $C_o{C}_1+C_1{C}_2+...+C_{n–1}{C}_n$

$(x+1{)}^n=C_o{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+...+C_n$

Step 2: Multiply $(1+x{)}^n$ and $(x+1{)}^n$

$(1+x{)}^{2n}=(C_o+C_{1}x+C_2{x}^2+...+C_n{x}^n)(C_o{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+...+C_n)$

Step 3: Compare the coefficients of $x^{n-1}$

$\left(\begin{array}{c}2n\\ n+1\end{array}\right)=$$C_o{C}_1+C_1{C}_2+...+C_{n-1}{C}_n$

$\frac{2n!}{(n-1)!(n+1)!}$ $=$$C_o{C}_1+C_1{C}_2+...+C_{n-1}{C}_n$

Hence, Option (A): $\frac{2n!}{(n-1)!(n+1)!}$ is the correct option.