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

(N/A) The $(r+1)^{\text{th}}$ term in the expansion of $(1+a)^{n}$ is given by $\binom{n}{r}a^{r}$. Thus,the coefficients of $a^{r-1}$,$a^{r}$,and $a^{r+1}$ are $\binom{n}{r-1}$,$\binom{n}{r}$,and $\binom{n}{r+1}$ respectively.
Since these coefficients are in arithmetic progression,we have $\binom{n}{r-1} + \binom{n}{r+1} = 2\binom{n}{r}$.
Expanding the binomial coefficients,we get:
$\frac{n!}{(r-1)!(n-r+1)!} + \frac{n!}{(r+1)!(n-r-1)!} = 2 \times \frac{n!}{r!(n-r)!}$
Dividing both sides by $n!$ and multiplying by $(r+1)!(n-r+1)!$,we simplify the expression:
$\frac{(r+1)r}{1} + \frac{(n-r+1)(n-r)}{1} = 2 \times \frac{(r+1)(n-r+1)}{1}$
$r^{2}+r + n^{2}-nr-nr+r^{2}+n-r = 2(nr-r^{2}+r+n-r+1)$
$n^{2} - 2nr + 2r^{2} + n = 2nr - 2r^{2} + 2n + 2$
$n^{2} - 4nr - n + 4r^{2} - 2 = 0$
$n^{2} - n(4r+1) + 4r^{2} - 2 = 0$.