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nth Term of A.P

if a0 , a...

Question

if $a_{0}, a_{1}, a_{2}$ ....be the coefficients in the expansion of $(1 + x + x$ $^{2})^{n}$ in ascending powers $f (x)$ , then prove that :
$(r + 1) a_{r + 1} = (n - r) a_{r} + (2 n - r + 1) a_{r - 1}, (0 < r < 2 n)$

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Solution

Differentiating both sides (1) w. r. t. x we get
$n (1 + x + x^{2})^{n - 1} (1 + 2 x) = \sum_{r = 0}^{2 n} r a_{r} x^{r - 1}$
Multiply both sides by $1 + x + x^{2}$ and replace $(1 + x + x^{2})^{n} b y \sum_{r = 0}^{2 n} a_{r} x^{r}$ we get
$n (2 x + 1) \sum_{r = 0}^{2 n} a_{r} x = (1 + x + x^{2})^{n - 1} \sum_{r = 0}^{2 n} r a_{r} x^{r - 1}$
Equating the coefficient of $x^{r}$ in both sides
$n [2 a_{r - 1} + a_{r}] = [(r + 1) a_{r + 1} + r a_{r} + (r - 1) a_{r - 1}]$
or $(r + 1) a_{r + 1} = (n - r) a_{r} + (2 n - r + 1) a_{r - 1}$
(0 < r < 2n )

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