If the coefficients of $a^{r-1}, a^{r}$ and $a^{r+1}$ in the expansion of $(1+a)^{n}$ are in arithmetic progression, prove that $n^{2}-n(4 r+1)+4 r^{2}-2=0$

The $(r+1)^{\text {th }}$ term in the expansion is ${\,^n}{C_r}{a^r}$. Thus it can be seen that $a^{r}$ occurs in the $(r+1)^{\text {th }}$ term, and its coefficient is ${\,^n}{C_r}$.

Hence the coefficients of $a^{r-1}, a^{r}$ and $a^{r+1}$ are ${\,^n}{C_{r - 1}},{\,^n}{C_r}$ and $^{n} C_{r+1},$ respectively. since these coefficients are in arithmetic progression, so we have, ${\,^n}{C_{r - 1}} + {\,^n}{C_{r + 1}} = 2.{\,^n}C,$ This gives

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

i.e.,      $\frac{1}{(r-1) !(n-r+1)(n-r)(n-r-1) !}+\frac{1}{(r+1)(r)(r-1) !(n-r-1) !}$

$=2 \times \frac{1}{r(r-1) !(n-r)(n-r-1) !}$

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

$=2 \times \frac{1}{(r-1) !(n-r-1) ![r(n-r)]}$

i.e.,    $\frac{1}{(n-r+1)(n-r)}+\frac{1}{r(r+1)}=\frac{2}{r(n-r)}$

or     $\frac{r(r+1)+(n-r)(n-r+1)}{(n-r)(n-r+1) r(r+1)}=\frac{2}{r(n-r)}$

or     $r(r+1)+(n-r)(n-r+1)=2(r+1)(n-r+1)$

or     $r^{2}+r+n^{2}-n r+n-n r+r^{2}-r=2\left(n r-r^{2}+r+n-r+1\right)$

or      $n^{2}-4 n r-n+4 r^{2}-2=0$

i.e.,     $n^{2}-n(4 r+1)+4 r^{2}-2=0$