Question

For all positive integers n, show that ²ⁿC_n + ²ⁿC_{n − 1} = (^{2n + 2}C_{n + 1}).

Answer 1

$$\begin{gathered} \mathrm{LHS}={}^{2n}{C}_n+{}^{2n}{C}_{n-1} \\ =\frac{\left(2n\right)!}{n!n!}+\frac{\left(2n\right)!}{\left(n-1\right)!\left(2n-n+1\right)!} \\ =\frac{\left(2n\right)!}{n!n!}+\frac{\left(2n\right)!}{\left(n-1\right)!\left(n+1\right)!} \\ =\frac{\left(2n\right)!}{n\left(n-1\right)!n!}+\frac{\left(2n\right)!}{\left(n-1\right)!\left(n+1\right)n!} \\ =\frac{\left(2n\right)!}{n!\left(n-1\right)!}\left[\frac{1}{n}+\frac{1}{n+1}\right] \\ =\frac{\left(2n\right)!}{n!\left(n-1\right)!}\left[\frac{2n+1}{n\left(n+1\right)}\right] \\ =\frac{\left(2n+1\right)!}{n!\left(n+1\right)!} \end{gathered}$$

$$\begin{gathered} \mathrm{RHS}=\frac{1}{2}{}^{2n+2}{C}_{n+1} \\ =\frac{1}{2}\left[\frac{\left(2n+2\right)!}{\left(n+1\right)!\left(2n+2-n-1\right)!}\right] \\ =\frac{1}{2}\left[\frac{\left(2n+2\right)!}{\left(n+1\right)!\left(n+1\right)!}\right] \\ =\frac{1}{2}\left[\frac{\left(2n+2\right)\left(2n+1\right)!}{\left(n+1\right)n!\left(n+1\right)!}\right] \\ =\frac{1}{2}\left[\frac{2\left(n+1\right)\left(2n+1\right)!}{\left(n+1\right)n!\left(n+1\right)!}\right] \\ =\frac{\left(2n+1\right)!}{n!\left(n+1\right)!} \end{gathered}$$

∴ LHS = RHS