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

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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 by thanvigandhi_1
 

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