$$\sum_{n=1}^{N}\frac {z^n}{2^n}$$
I attempted to use the method of difference but it didn't work.
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It is $$\sum_{n=1}^{N}(z/2)^n.$$ – Math Lover May 11 '18 at 17:38
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1Why rush for posting an answer before searching for duplicates? – GNUSupporter 8964民主女神 地下教會 May 11 '18 at 17:45
2 Answers
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Note that you can write $$1+x+x^2+...+x^N=(1+x+x^2+...+x^N)(1-x)/(1-x)=(1-x^{N+1})/(1-x)$$ if $x\ne 1$
Andrei
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Use the formula for the sum of consecutive terms of a geometric progression: $$\frac z2\,\frac{\Bigl(\dfrac z2\Bigr)^{\!N}-1}{\dfrac z2-1}=\frac{z\Bigl(z^N-2^N\Bigr)}{2^N(z-2)}\qquad(z\ne 2)$$
Bernard
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