# Prove the following by using the principle of mathematical induction for all $n in N: 1.3+2.3^{2}+3.3^{3}+ldots+n .3^{n}$

Question:

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

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): 1.3+2.3^{2}+3.3^{3}+\ldots+n 3^{n}=\frac{(2 n-1) 3^{n+1}+3}{4}$

For $n=1$, we have

$P(1): 1.3=3=\frac{(2.1-1) 3^{1+1}+3}{4}=\frac{3^{2}+3}{4}=\frac{12}{4}=3$, which is true.

Let $P(k)$ be true for some positive integer $k$, i.e.,

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

We shall now prove that P(k + 1) is true.

Consider

$1.3+2.3^{2}+3.3^{3}+\ldots+k 3^{k}+(k+1) 3^{k+1}$

$=\left(1.3+2.3^{2}+3.3^{3}+\ldots+k .3^{k}\right)+(k+1) 3^{k+1}$

$=\frac{(2 k-1) 3^{k+1}+3}{4}+(k+1) 3^{k+1} \quad$ [Using] (i)

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

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

$=\frac{3^{k+1}\{6 k+3\}+3}{4}$

$=\frac{3^{k+1} \cdot 3\{2 k+1\}+3}{4}$

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

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

Thus, $\mathrm{P}(k+1)$ is true whenever $\mathrm{P}(k)$ is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.