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

Let the given statement be $P(n)$, i.e.,
$P(n): 1.3+2.3^2+3.3^3+...+n.3^n=\large\frac{(2n-1)3^{n+1}+3}{4}$
For $n = 1$, we have
$P(1):1.3 = 3= \large\frac{(2.1-1)3^{1+1}+3}{4}$$=\large\frac{3^2+3}{4}$$=\large\frac{12}{4}$$=3, which is true. Let P(k) be true for some positive integer k, i.e., 1.3+2.3^2+3.3^3+...+k3^k=\large\frac{(2k-1)3^{k+1}+3}{4}--------(i) We shall now prove that P(k+1) is true. Consider 1.3+2.3^2+3.3^3+...+k3^k+(k+1)3^{k+1} = (1.3+2.3^2+3.3^3+...+k3^k)+(k+1)3^{k+1} = \large\frac{(2k-1)3^{k+1}+3}{4}$$+(k+1)3^{k+1}\qquad$ [ Using (i) ]
$= \large\frac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}$
$= \large\frac{3^{k+1} \{2k-1+4(k+1) \}+3}{4}$
$= \large\frac{3^{k+1} \{6k+3 \}+3}{4}$
$= \large\frac{3^{k+1}.3 \{2k+1 \}+3}{4}$
$= \large\frac{3^{(k+1)+1} \{2k+1 \}+3}{4}$
$= \large\frac{ \{2(k+1)-1 \} 3^{(k+1)+1}+3}{4}$
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.
answered Apr 29, 2014