Question 267871
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{{{sum(3^i,i=1,n) = 3(3^n-1)/2)

}}}

It's true for n=1, because:

{{{sum(3^i,i=1,1) = 3^1 = 3 = 3*2/2= (3(3-1))/2=(3(3^1-1))/2

}}}


Assume true for any k such that {{{1<=n<=k}}}

We want to prove that

{{{sum(3^i,i=1,k+1) = 3(3^(k+1)-1)/2)}}}

Write that as the sum to k (for which we assume the formula holds,
plus the (k+1)st term or {{{3^(k+1)}}}

{{{sum(3^i,i=1,k+1) = (sum(3^i,i=1,k))+3^(k+1) = 3(3^k-1)/2 + 3^(k+1) =

(3*3^k-3)/2 + (2*3^(k+1))/2=(3^(k+1)-3)/2+(2*3^(k+1))/2 =

(3^(k+1)-3+2*3^(k+1))/2 = (3*3^(k+1)-3)/2=(3(3^(k+1)-1))/2}}} 

Edwin</pre>