Question

Prove the following by using the principle of mathematical induction for all $n\in N:1.3+2.3^2+3.3^3+.....+n.3{}^n=\frac{(2n-1)3^{n+1}+3}{4}$

Answer 1

Let the given statement be $P\left(n\right)$ i.e.,
$1.3+2.3^2+3.3^3+........n.3^n=\frac{(2n-1)3^{n+1}+3}{4}$
$P\left(n\right)$:
For $n=1$, we have
$P\left(1\right):1.3=3=\frac{(2.1-1)3^{1+1}+3}{4}=\frac{3^2+3}{4}=\frac{12}{4}=\mathrm{3........}\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider
$1.3+2.3^2+3.3^3+.......k.3^k+(k+1)3^{k+1}=(1.3+2.3^2+3.3^3+.....+k.3^k+(k+1)3^{k+1}$
$=\frac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k+1}$ [Using $\left(i\right)$]
$=\frac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}=\frac{3^{k+1}\{2k-1+4(k+1\left)\right\}+3}{4}$
$=\frac{3^{(k+1)+1}\{2k+1\}+3}{4}$
$=\frac{\left\{2\right(k+1)-1\}{3}^{(k+1)+1}+3}{4}$
Thus $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction. statement $P\left(n\right)$ is true for all natural numbers i.e., $n$.