Question

Prove that the coefficient of (r+1)th term in the expansion of $(1+x{)}^{n+1}$ is equal to the sum of the coefficients of rth and (r+1)th terms in the expansion of $(1+x{)}^n.$

Answer 1

Now, coefficient of (r+1)th term in the expansion of $(4+x{)}^{n+1}{=}^{n+1}{C}_{r+1-1}$

$={}^{n+1}{C}_r$ and, coefficient of $r^{th}$ term in $(1+x{)}^n$ +coefficient of $(r+1{)}^{th}$ term in $(1+x{)}^n$

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

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

$=\frac{n!}{\{n-(r-1\left)\right\}(r-1)!}+\frac{n!}{(n-r)!r!}$

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

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

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

$=\frac{n!}{(n-r)!(r-1)!}[\frac{1}{n-r+1}+\frac{1}{r}]$

$=\frac{n!}{(n-r)!(r-1)!}\left[\frac{r+n-r+1}{(n-r+1)r}\right]$

$=\frac{n!}{(n-r)!(r-1)!}\left[\frac{n+1}{(n-r+1)r}\right]$

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

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

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

$={}^{n+1}{C}_r$

$\therefore {}^{n+1}{C}_r={}^n{C}_{r-1}+{}^n{C}_r$

The coefficient of (r+1)th term in the expansion of $(1+x{)}^{n+1}$ is equal to the sum of the coefficient of rth and $(r+1{)}^{th}$ terms in the expansion of $(1+x{)}^n.$