Question

Prove for $n\in N$.
$1.3+{2.3}^2+{3.3}^3+...+n{.3}^n=\frac{\left(2n-1\right)3^{n+1}+3}{4}$

Answer 1

Given

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

For $n=1$

L.H.S $={\mathrm{1.3.}}^1=3$

R.H.S$=\frac{(R-1-1){3.}^{1+1}+3}{4}$

$=\frac{{1.3}^2+3}{4}=\frac{9+3}{4}=\frac{12}{4}=3$

$\therefore$ L.H.S=R.H.S

True for $n=1$

Let $n=k$

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

For $n=k+1$

$1.3+{2.3}^2+...\dots \dots +k{.3}^k+(k+1){.3}^{k+1}$

Upto k. put above value

$=\frac{(2k-1){.3}^{k+1}+3}{4}+(k+1){.3}^{k+1}$

$=\frac{(2k-1){.3}^{k+1}+3+4(k+1){.3}^{k+1}}{4}$

$=\frac{3^{k+1}(2k-1+4k+4)}{4}$

$=\frac{{3.}^{k+1}(6k+3)}{4}=\frac{3^{k+1}.3(2k+1)+3}{4}$

$=\frac{3^{k+1}\left(2\right(k+1)-1)+3}{4}$

Hence true for $n=k+1$.