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Question Number 183726 by LEKOUMA last updated on 29/Dec/22

Prove that Σ_(k=0) ^(n−1) 2^(n−1) =n×2^(n−1)

Provethat∑n−1k=02n−1=n×2n−1

Commented by JDamian last updated on 29/Dec/22

seriously?

Answered by Vynho last updated on 29/Dec/22

it′s the sum of 2^(n−1) at [(n−1)+1] time

it′sthesumof2n−1at[(n−1)+1]time

Answered by Ar Brandon last updated on 29/Dec/22

Σ_(k=0) ^(n−1) 2^(n−1) =2^(n−1) Σ_(k=0) ^(n−1) (1)=2^(n−1) ×n=n×2^(n−1)

∑n−1k=02n−1=2n−1∑n−1k=0(1)=2n−1×n=n×2n−1

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