Question

$1.3+{2.3}^2+{3.3}^3+\cdots +n.3^n=\frac{(2n-1)3^{n+1}+3}{4}$

Answer 1

Let $P\left(n\right)=1.3+{2.3}^2+{3.3}^3+\cdots +n.3^n=\frac{(2n-1)3^{n+1}+3}{4}$
For n = 1
$P\left(1\right)=1.3=\frac{(2\times 1-1)3^{1+1+3}}{4}\Rightarrow 3=\frac{9+3}{4}\Rightarrow 3=3\therefore P\left(1\right)$
Let p(n) be true for n = k
$\therefore P\left(k\right)=1.3+2{.3}^2+{3.3}^3+\cdots +k.3^k=\frac{(2k-1)k^{k+1}+3}{4}\text{for}n=k+1P(k+1)=1.3+{2.3}^2+{3.3}^3+\cdots +k.3^K+(k+1).3^{k+1}=\frac{(2k-1)3^{k+1}+3}{4}+(k+1).3^{k+1}=\frac{(2k-1){.3}^{k+1}}{4}+\frac{3}{4}+(k+1)3^{k+1}=3^{k+1}[\frac{2k-1}{4}+k+1]+\frac{3}{4}=3^{k+1}\left[\frac{2k-1+4k+4}{4}\right]+\frac{3}{4}=3^{k+1}\left[\frac{6k+3}{4}\right]+\frac{3}{4}=\frac{3^{k+1}.3(2k+1)}{4}+\frac{3}{4}=\frac{(2k+1)3^{k+2}+3}{4}$
$\therefore$ P(k + 1) is true
Thus P (k ) is true $\Rightarrow P(k+1)$ is true Hence by principle fo mathematical induction