If the coefficient of $r^{t h}, (r + 1)^{t h}$ and $(r + 2)^{t h}$ terms in the expansion of $(1 + x)^{n}$ are in A.P., then show that $n^{2} - (4 r + 1) n + 4 r^{2} - 2 = 0$ .

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Solution

The general term of the binomial expansion $(1 + x)^{n}$ can be written as
$T_{r + 1} =^{n} C_{r} x^{r}$
Therefore the coefficients of $r^{t h}$ , $(r + 1)^{t h}$ and $(r + 2)^{t h}$ terms will be
$^{n} C_{r - 1},^{n} C_{r},^{n} C_{r + 1}$
In an A.P the middle term is the mean/average of the first and last terms, hence
$^{n} C_{r} = \frac{^{n} C_{r - 1} +^{n} C_{r + 1}}{2}$
$2^{n} C_{r} =^{n} C_{r - 1} +^{n} C_{r + 1}$
$\frac{2 (n!)}{(n - r)! r!} = \frac{(n!)}{(n - r + 1)! (r - 1)!} + \frac{(n!)}{(n - 1 - r)! (r + 1)!}$
$\frac{2}{(n - r)! r} = \frac{1}{(n - r + 1)!} + \frac{1}{(n - 1 - r)! (r + 1) r}$
$\frac{2}{(n - r) (r)} = \frac{1}{(n - r) (n - r + 1)} + \frac{1}{(r + 1) . r}$
$\frac{2}{(n - r) (r)} - \frac{1}{(n - r) (n - r + 1)} = \frac{1}{(r + 1) . r}$
$\frac{2 n - 2 r + 2 - r}{r . (n - r + 1)} = \frac{(n - r)}{(r + 1) (r)}$
$\frac{2 n - 3 r + 2}{(n - r + 1)} = \frac{(n - r)}{(r + 1)}$
$(2 n - 3 r + 2) (r + 1) = (n - r) (n - r + 1)$

$2 n r - 3 r^{2} + 2 r + 2 n - 3 r + 2 = n^{2} - n r + n - n r + r^{2} - r$
$- 3 r^{2} - r + 2 n r + 2 n + 2 = n^{2} - 2 n r + n + r^{2} - r$
$4 r^{2} - 4 n r - n + n^{2} - 2 = 0$

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

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