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Q.
If the coefficients of $a^{r-1}, a^r$ and $a^{r+1}$ in the expansion of $(1+a)^n$ are in arithmetic progression, then $n^2-n(4 r+1)+4 r^2-2$ is equal to
Binomial Theorem
Solution:
The $(r+1)^{\text {th }}$ term in the expansion is ${ }^n C_r a^r$. Thus, it can be seen that $a^{\prime}$ 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 ${ }^n C_{r-1}+{ }^n C_{r+1}=2 .{ }^n C_r$. 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$