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Quick GP sum

Authors
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    Name
    Vu Hung
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Problem Statement

Find

1+12+14+18++1128.1+\frac12+\frac14+\frac18+\cdots+\frac1{128}.

Hints

This is a finite GP with a=1a=1, r=12r=\frac12, and 88 terms.


Solutions

S8=1(12)8112=2(11256)=255128.S_8=\frac{1-(\frac12)^8}{1-\frac12}=2\left(1-\frac1{256}\right)=\frac{255}{128}.

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