Question

Prove by PMI 3.2^2 +3^2.2^3 + 3^3.2^4 +----+ 3^n.2^n+1 = 12/5(6^n -1)

Answer 1

We can see that, when $n=1$, $LHS=3.2^2=12\\ RHS=\frac{12(6^1-1)}{5} =12$

The given sum is true for $n=1$

Let the given sum is true for $n=m$. Then

$S(m)=3.2^2+3^2.2^3+...+3^m.2^{m+1}=\frac{12(6^m-1)}{5}$

Now ,.

$S(m+1)=3.2^2+3^2.2^3+...+3^m.2^{m+1}+3^{m+1}.2^{m+2}=\frac{12(6^m-1)}{5} +3^{m+1}.2^{m+2}\\ S(m+1)=\frac{12(6^m-1)}{5} +12.6^{m}\\ S(m+1)=12\frac{(6^m-1+5.6^m)}{5} \\ S(m+1)=12\frac{(6^m-1+(6-1).6^m)}{5} \\ S(m+1)=12\frac{(6^{m+1}-1)}{5}$

Which is true. So if the sum is true for $n=m$ , then the sum is true for $n=m+1$. Thus by the principle of mathematical induction, the proof is complete.