Subjective Type

Let there be $$a_1, a_2, a_3, \dots, a_n$$ terms in G.P. whose common ratio is r. Let $$S_k$$ denote the sum of first k terms of this G.P. Prove that $$S_{m-1} S_m = \frac{r+1}{r} \displaystyle \sum_{i < j}^{m} a_ia_j$$.

Solution

We have,

$$(a_1 + a_2 + \dots + a_m)^2 = a_1^2 + a_2^2 + \dots + a_m^2 + 2(a_1a_2 + a_2a_3 + \dots)$$

or

$$\displaystyle \Bigg[ \frac{a_1 (1 - r^m)}{1-r} \Bigg]^2 = \frac{a_1^2 (1 - r^{2m})}{1-r^2} + 2 \sum_{i
$$\displaystyle \Rightarrow 2 \sum_{i
$$ = \frac{2a_1^2}{(1-r)^2(1 + r)} [r - r^m - r^{m+1} + r^{2m}]$$

$$ = \frac{2r}{1+r} \Bigg\{ a_1 \times \frac{(1-r^{m-1})}{1-r} \Bigg\} \Bigg\{ \frac{a_1(1 - r^m)}{1-r} \Bigg\}$$

$$\displaystyle \Rightarrow \frac{r+1}{r} \sum_{i

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