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Polynomial and Its General Form

The set a 1...

Question

The set ${a_{1}, a_{2}, . . . . a_{n}}$ form a $G . P .$ with first term as $a$ and common ratio $r$ . Prove that $\sum a_{i} a_{j} = \frac{a^{2} r ({1 - r}^{n - 1}) ({1 - r}^{n})}{{(1 - r)}^{2} (1 + r)}$

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Solution

$\sum a_{i} = S = \frac{a (1 - r^{n})}{1 - r}$ ...... $(1)$
$\sum a_{i}^{2} = a^{2} + a^{2} r^{2} + a^{2} r^{4} + . . . . . . = \frac{a^{2} (1 - r^{2 n})}{1 - r^{2}}$ ...... $(2)$
Now ${(\sum a_{i})}^{2} = \sum a_{i}^{2} + 2 \sum a_{i} a_{j}$
$∴ 2 \sum a_{i} a_{j} = {(\sum a_{i})}^{2} - \sum a_{i}^{2}$
$= \frac{a^{2} {(1 - r^{n})}^{2}}{{(1 - r)}^{2}} - \frac{a^{2} (1 - r^{2 n})}{1 - r^{2}}$ by $(1)$ and $(2)$
$= \frac{a^{2} (1 - r^{n})}{(1 - r)} [\frac{1 - r^{n}}{1 - r} - \frac{1 - {+ r}^{n}}{1 + r}]$
$\sum a_{i} a_{j} = \frac{1}{2} \frac{a^{2} r (1 - r^{n - 1}) (1 - r^{n})}{{(1 - r)}^{2} (1 + r)}$

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