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

$$\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{to}\mathrm{n}\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{series}$$$$1+\frac{\mathrm{x}}{\mathrm{a}}(1+\mathrm{x})+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}(1+\mathrm{x}+{\mathrm{x}}^2)+\frac{{\mathrm{x}}^3}{{\mathrm{a}}^3}(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3)+\dots$$$$$$

Answered by mindispower last updated on 30/Sep/20

$$=\frac{1}{x-1}[+\frac{x(x^2-1)}{a}+\frac{x^2}{a^2}(x^3-1)+......3$$$$=\frac{1}{x-1}[1+\sum _{k⩾1}\frac{x^{2k+1}}{a^k}-\sum _{k⩾1}\frac{x^k}{a^k}]$$

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

$$\mathrm{S}=1+\frac{\mathrm{x}}{\mathrm{a}}(1+\mathrm{x})+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}(1+\mathrm{x}+{\mathrm{x}}^2)+....+\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{a}}^{\mathrm{n}-1}}.\frac{1-{\mathrm{x}}^{\mathrm{n}}}{1-\mathrm{x}}$$$$-\frac{\mathrm{x}}{\mathrm{a}}\mathrm{S}=-\frac{\mathrm{x}}{\mathrm{a}}-\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}(1+\mathrm{x})+.......-\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{a}}^{\mathrm{n}-1}}.\frac{1-{\mathrm{x}}^{\mathrm{n}-1}}{1-\mathrm{x}}-\frac{\mathrm{x}}{\mathrm{a}}.\frac{1-{\mathrm{x}}^{\mathrm{n}}}{1-\mathrm{x}}$$$$(1-\frac{\mathrm{x}}{\mathrm{a}})\mathrm{S}=1+\frac{\mathrm{x}}{\mathrm{a}}\mathrm{x}+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}{\mathrm{x}}^2+\frac{{\mathrm{x}}^3}{{\mathrm{a}}^3}.{\mathrm{x}}^3+....\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{a}}^{\mathrm{n}-1}}.{\mathrm{x}}^{\mathrm{n}-1}-\frac{\mathrm{x}}{\mathrm{a}}.\frac{1-{\mathrm{x}}^{\mathrm{n}}}{1-\mathrm{x}}$$$$(1-\frac{\mathrm{x}}{\mathrm{a}})\mathrm{S}=\frac{1-{\left(\frac{{\mathrm{x}}^2}{\mathrm{a}}\right)}^{\mathrm{n}}}{1-\frac{{\mathrm{x}}^2}{\mathrm{a}}}-\frac{\mathrm{x}}{\mathrm{a}}.\frac{1-{\mathrm{x}}^{\mathrm{n}}}{1-\mathrm{x}}$$$$(1-\frac{\mathrm{x}}{\mathrm{a}})\mathrm{S}=\frac{1}{{\mathrm{a}}^{\mathrm{n}-1}}\frac{{\mathrm{a}}^{\mathrm{n}}-{\mathrm{x}}^{2\mathrm{n}}}{\mathrm{a}-{\mathrm{x}}^2}-\frac{\mathrm{x}}{\mathrm{a}}.\frac{1-{\mathrm{x}}^{\mathrm{n}}}{1-\mathrm{x}}$$$$\mathrm{S}=\frac{1}{{\mathrm{a}}^{\mathrm{n}-2}(\mathrm{a}-\mathrm{x})}.\frac{{\mathrm{a}}^{\mathrm{n}}-{\mathrm{x}}^{2\mathrm{n}}}{\mathrm{a}-{\mathrm{x}}^2}-\frac{\mathrm{x}}{{\mathrm{a}}^{\mathrm{n}-1}(\mathrm{a}-\mathrm{x})}.\frac{1-{\mathrm{x}}^{\mathrm{n}}}{1-\mathrm{x}}$$

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