Prove-that-k-0-n-1-2-n-1-n-2-n-1-

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

P r o v e t h a t \underset{k = 0}{\sum^{n - 1}} 2^{n - 1} = n \times 2^{n - 1}

Commented by JDamian last updated on 29/Dec/22

seriously?

Answered by Vynho last updated on 29/Dec/22

{i t}^{'} s t h e s u m o f 2^{n - 1} a t [(n - 1) + 1] t i m e

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

\underset{k = 0}{\sum^{n - 1}} 2^{n - 1} = 2^{n - 1} \underset{k = 0}{\sum^{n - 1}} (1) = 2^{n - 1} \times n = n \times 2^{n - 1}