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#