M.I question....

When n = 1,
RHS = 1/4(1+(2*1-1)*3^1) = 1 = LHS
The proposition is true for n = 1.

Assume it is true for n = k,
ie. 1+2*3+3*3^2+4*3^3+...+k(3^<k-1>)=1/4[1+(2k-1)3^k] for all positive integers k.

So we are going to prove LHS = RHS for n = k+1.
(Remark: we are going to prove LHS = 1/4{1+[2(k+1)-1]*3^<k+1>})
LHS
= 1+2*3+3*3^2+4*3^3+...+k(3^<k-1>)+(k+1)*3^k
= 1/4[1+(2k-1)*3^k]+(k+1)*3^k (by assumption)
= 1/4[1+(2k-1)*3^k+(4k+4)*3^k]
= 1/4[1+(6k+3)*3^k]
= 1/4{1+[6(k+1)-3]*3^k}
= 1/4{1+[2(k+1)-1]*3^<k+1>}
= RHS

By the first principle of mathematical induction,
it is true for all positive integers n.

As follow AS~~~

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