Show by induction that sum_n(r*3^(r-1))=1/4+(3^n/4)*(2n-1) for n>0
Base Case n=1sum_1(r3^(r-1))=13^0=1=1/4+(3^1/4)(2-1)=1/4+3/4=1
Assume true for n=ki.e. sum_k(r3^(r-1))=(3^k/4)(2k-1)
then for n=k+1sum_(k+1)(r3^(r-1))= sum_k(r3^(r-1))+(k+1)3^k =(3^k/4)(2k-1)+(k+1)3^k =(3^k/4)(2k-1+4k+4)=(3^k/4)(6k+3)=(3^(k+1)/4)(2k+1)=(3^(k+1)/4)(2*(k+1)-1)
Therefore by inductionsum_n(r3^(r-1))=1/4+(3^n/4)(2n-1) for n>0
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