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Question Number 133146 by shaker last updated on 19/Feb/21

Commented by Dwaipayan Shikari last updated on 19/Feb/21

$$\frac{1-a^{n+1}}{1-a}$$

Answered by mathmax by abdo last updated on 19/Feb/21

$$\mathrm{if}\mathrm{a}\ne 1\mathrm{we}\mathrm{have}1+\mathrm{a}+{\mathrm{a}}^2+...+{\mathrm{a}}^{\mathrm{n}}=\frac{1-{\mathrm{a}}^{\mathrm{n}+1}}{1-\mathrm{a}}$$$$\mathrm{if}\mathrm{a}=1\mathrm{we}\mathrm{get}1+\mathrm{a}+{\mathrm{a}}^2+...+{\mathrm{a}}^{\mathrm{n}}=\mathrm{n}+1$$

Answered by physicstutes last updated on 19/Feb/21

$$a=1,r=a$$$${\mathrm{S}}_n=\frac{1(1-a^{n+1})}{1-a}$$

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