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Q.
For all $n \in N , 1.3+2.3^{2}+3.3^{3}+\ldots . .+ n .3^{ n }$ is equal to
Principle of Mathematical Induction
Solution:
Let the statement $P ( n )$ be defined as
$P(n)=1.3+2.3^{2}+3.3^{3}+\ldots . .+n .3^{n}=\frac{(2 n-1) 3^{n+1}+3}{4}$
Step $I :$ For $n =1$
$P (1): 1.3=\frac{(2.1-1) 3^{1+1}+3}{4}=\frac{3^{2}+3}{4}$
$=\frac{9+3}{4}=\frac{12}{4}=3=1.3$, which is true.
Step II : Let it is true for $n = k$
i.e. $1.3+2.3^{2}+3.3^{3}+\ldots \ldots+ k .3^{ k }$
$=\frac{(2 k-1) 3^{k+1}+3}{4} \dots$(i)
Step III $:$ For $n = k +1$
$\left(1.3+2.3^{2}+3.3^{3}+\ldots . .+ k .3^{ k }\right)+( k +1) 3^{ k +1}$
$=\frac{(2 k -1) 3^{ k +1}+3}{4}+( k +1) 3^{ k +1}$
[Using equation (i)]
$=\frac{(2 k-1) 3^{k+1}+3+4(k+1) 3^{k+1}}{4}$
$=\frac{3^{ k +1}(2 k -1+4 k +4)+3}{4}$
[taking $3^{ k +1}$ common in first and last term of numerator part $]$
$=\frac{3^{ k +1}(6 k +3)+3}{4}$
$=\frac{3^{ k +1} \cdot 3(2 k +1)+3}{4}$
[taking 3 common in first term of numerator part]
$=\frac{3^{( k +1)+1}[2 k +2-1]+3}{4}$
$=\frac{[2( k +1)-1] 3^{( k +1)+1}+3}{4}$
Therefore, $P ( k +1)$ is true when $P ( k )$ is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers $n$