Please can you prove this?

Question

$1+3+3^2+⋯⋯⋯⋯3^(n-1)= (3^n-1)/2$

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Answer 1 · 2018-04-22T08:31:40

See explanation

We want to prove

$1+3+3^2+...+3^(n-1)=(3^n-1)/2$

Let us call

$S=1+3+3^2+...+3^(n-1)$

Multiply both sides by 3

$3S=3+3^2+...+3^(n-1)+3^n$

So by the definition of $S$

$3S=(S-1)+3^n$

$=>2S=3^n-1$

$=>S=(3^n-1)/2$

Or

$1+3+3^2+...+3^(n-1)=(3^n-1)/2$