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I am not able to solve this sigma question. Please anybody solve this question by steps. $$\sum_{R=1}^N\left(\frac13\right)^{R-1}$$

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closed as off-topic by Saad, TheGeekGreek, José Carlos Santos, Claude Leibovici, Gibbs Jul 18 '18 at 9:16

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    $\begingroup$ That's a geometric progression. Bad luck at Wimbledon! $\endgroup$ – Lord Shark the Unknown Jul 18 '18 at 6:00
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Let $r=\frac13$, and $$S_N=\sum_{R=1}^N\left(\frac13\right)^{R-1}=\sum_{R=1}^Nr^{R-1}$$

You have $$rS_N=r\sum_{R=1}^Nr^{R-1}=\sum_{R=1}^Nr^R$$ Then $$S(r-1)=rS-S=\sum_{R=1}^Nr^R-\sum_{R=1}^Nr^{R-1}=\cdots=r^N-r^0$$ So $$S=\dfrac{1-r^N}{1-r}=\frac32\left(1-\frac1{3^N}\right)$$

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