Question

$C_3/4+C_5/6+C_7/8+.....=$

• A. $\frac{2^{n+1}-n^2-n-2}{2(n+1)}$
• B. $\frac{2^{n+2}+n^2+n-2)}{n+1}$
• C. $\frac{(3^{n+2}-n^2-n-2)}{n+1}$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{2^{n+1}-n^2-n-2}{2(n+1)}$

$\frac{C_3}{4}+\frac{C_5}{6}+\frac{C_7}{8}+......=\frac{(n!)}{(n-3)!3!4}+\frac{(n!)}{(n-5)!5!6}+\frac{(n!)}{(n-7)!7!8}+...=\frac{(n!)}{(n-3)!4!}+\frac{(n!)}{(n-5)!6!}+\frac{(n!)}{(n-7)!8!}+...=\frac{1}{(n+1)}(\frac{(n+1)!}{(n+1-4)!4!}+\frac{(n+1)!}{(n+1-6)!6!}+\frac{(n+1)!}{(n+1-8)!8!}+...)=\frac{1}{(n+1)}(\frac{(n+1)!}{(n+1-4)!4!}+\frac{(n+1)!}{(n+1-6)!6!}+\frac{(n+1)!}{(n+1-8)!8!}+...)=\frac{1}{(n+1)}(\left(\begin{array}{c}n+1\\ 4\end{array}\right)+\left(\begin{array}{c}n+1\\ 6\end{array}\right)+\left(\begin{array}{c}n+1\\ 8\end{array}\right)+...)=\frac{1}{(n+1)}(\left(\begin{array}{c}n+1\\ 0\end{array}\right)+\left(\begin{array}{c}n+1\\ 2\end{array}\right)+\left(\begin{array}{c}n+1\\ 4\end{array}\right)+\left(\begin{array}{c}n+1\\ 6\end{array}\right)+\left(\begin{array}{c}n+1\\ 8\end{array}\right)+...-\left(\begin{array}{c}n+1\\ 0\end{array}\right)-\left(\begin{array}{c}n+1\\ 2\end{array}\right))=\frac{1}{(n+1)}(2^{n+1-1}-1-\frac{n(n+1)}{2})(\because \left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)+\left(\begin{array}{c}n\\ 4\end{array}\right)+...=2^{n-1})=\frac{2^n-1}{n+1}-\frac{n}{2}=\frac{2^{n+1}-2-n-n^2}{2(n+1)}$