Find-the-sum-to-n-terms-of-the-series-1-x-a-1-x-x-2-a-2-1-x-x-2-x-3-a-3-1-x-x-2-x-3-

Question Number 116023 by aye48 last updated on 30/Sep/20

Find the sum to n terms of the series

1 + \frac{x}{a} (1 + x) + \frac{x^{2}}{a^{2}} (1 + x + x^{2}) + \frac{x^{3}}{a^{3}} (1 + x + x^{2} + x^{3}) + \dots

Answered by mindispower last updated on 30/Sep/20

= \frac{1}{x - 1} [+ \frac{x (x^{2} - 1)}{a} + \frac{x^{2}}{a^{2}} (x^{3} - 1) + \dots \dots 3

= \frac{1}{x - 1} [1 + \sum_{k ⩾ 1} \frac{x^{2 k + 1}}{a^{k}} - \sum_{k ⩾ 1} \frac{x^{k}}{a^{k}}]

Answered by Dwaipayan Shikari last updated on 30/Sep/20

S = 1 + \frac{x}{a} (1 + x) + \frac{x^{2}}{a^{2}} (1 + x + x^{2}) + \dots . + \frac{x^{n - 1}}{a^{n - 1}} . \frac{1 - x^{n}}{1 - x}

- \frac{x}{a} S = - \frac{x}{a} - \frac{x^{2}}{a^{2}} (1 + x) + \dots \dots . - \frac{x^{n - 1}}{a^{n - 1}} . \frac{1 - x^{n - 1}}{1 - x} - \frac{x}{a} . \frac{1 - x^{n}}{1 - x}

(1 - \frac{x}{a}) S = 1 + \frac{x}{a} x + \frac{x^{2}}{a^{2}} x^{2} + \frac{x^{3}}{a^{3}} . x^{3} + \dots . \frac{x^{n - 1}}{a^{n - 1}} . x^{n - 1} - \frac{x}{a} . \frac{1 - x^{n}}{1 - x}

(1 - \frac{x}{a}) S = \frac{1 - {(\frac{x^{2}}{a})}^{n}}{1 - \frac{x^{2}}{a}} - \frac{x}{a} . \frac{1 - x^{n}}{1 - x}

(1 - \frac{x}{a}) S = \frac{1}{a^{n - 1}} \frac{a^{n} - x^{2 n}}{a - x^{2}} - \frac{x}{a} . \frac{1 - x^{n}}{1 - x}

S = \frac{1}{a^{n - 2} (a - x)} . \frac{a^{n} - x^{2 n}}{a - x^{2}} - \frac{x}{a^{n - 1} (a - x)} . \frac{1 - x^{n}}{1 - x}