Question

How to prove ${{}^{n}C}_r+{{}^{n}C}_{r-1}={{}^{n+1}C}_r$

Answer 1

Prove the given expression :

$\begin{array}{rcl}L\mathit{.}H\mathit{.}S& =& {{}^{n}C}_r+{{}^{n}C}_{r-1}\\ & =& \frac{n!}{r!(n-r)!}+\frac{n!}{\left(r-1\right)!(n-\left(r-1\right))!}[\because {{}^{n}C}_r=\frac{n!}{r!(n-r)!}]\\ & =& n!\left[\frac{1}{r!(n-r)!}+\frac{1}{\left(r-1\right)!(n-r+1)!}\right]\\ & =& \frac{n!}{\left(r-1\right)!}\left[\frac{1}{r(n-r)!}+\frac{1}{(n-r+1)!}\right]\\ & =& \frac{n!}{\left(r-1\right)!(n-r)!}\left[\frac{1}{r}+\frac{1}{n-r+1}\right]\\ & =& \frac{n!}{\left(r-1\right)!(n-r)!}\left[\frac{n-r+1+r}{(n-r+1)r}\right]\\ & =& \frac{n!}{\left(r-1\right)!(n-r)!}\left[\frac{n+1}{(n-r+1)r}\right]\\ & =& \frac{n!(n+1)}{\left\{r\left(r-1\right)!\right\}(n-r)!}\left[\frac{1}{(n-r+1)}\right]\left[\because r\left(r-1\right)!=r!\right]\\ & =& \frac{(n+1)!}{r!(n-r)!}·\frac{1}{(n+1-r)}\left[\because n!\left(n+1\right)!=\left(n+1\right)!\right]\\ & =& \frac{(n+1)!}{r!(n+1-r)!}\\ & =& {{}^{n+1}C}_r[\because {{}^{n}C}_r=\frac{n!}{r!(n-r)!}]\\ & =& R\mathit{.}H\mathit{.}S\end{array}$

So, $L.HS=R.HS$

Hence,${{}^{n}C}_r+{{}^{n}C}_{r-1}={{}^{n+1}C}_r$