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

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Solution

We know that
$(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + \dots +^{n} C_{r - 1} x^{r - 1} +^{n} C_{1} x^{r} +^{n} C_{r + 1} X^{r + 1} + \dots$
So, the coefficients of $x^{r - 1} x^{r} a n d x^{r + 1}$ in the given expansion are $^{n} C_{r - 1},^{n} C_{r},$ and $^{n} C_{r + 1}$ respectively.
It is given that $^{n} C_{r - 1^{'}}^{n} C_{r}, {and}^{n} C_{r + 1} r+ 1 are in AP. ∴ 2 \times^{n} C_{r} =^{n} C_{r - 1} +^{n} C_{r + 1} \Rightarrow \frac{^{n} C_{r - 1}}{^{n} C_{r}} + \frac{^{N} C_{r + 1}}{^{n} C_{r}} = 2 \Rightarrow \frac{n!}{(r - 1)! . (n - r + 1)!} \times \frac{(r!) . (n - r)!}{n!} + \frac{n!}{(r + 1)! . (n - r + 1)!} \times \frac{(r!) . (n - r)!}{n!} = 2 \Rightarrow \frac{r}{(n - r + 1)} + \frac{(n - r)}{r + 1} = 2 [∵ (n - r + 1)! = (n - r + 1) . {(n - r)! r!} = r . {(r - 1)!}] \Rightarrow r (r + 1) + (n - r) (n - r + 1) = 2 (r + 1) (r + 1) (n - r + 1) \Rightarrow n^{2} - n (4 r + 1) + 4 r^{2} = 2, hence , n^{2} - n (4 r + 1) + 4 r^{2} - 2 = 0$

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