Question

For all $n\in N,1.3+2.3^2+3.3^3+\dots ..+n.3^n$ is equal to

• A. $\frac{(2n+1)3^{n+1}+3}{4}$
• B. $\frac{(2n-1)3^{n+1}+3}{4}$
• C. $\frac{(2n+1)3^n+3}{4}$
• D. $\frac{(2n-1)3^{n+1}+1}{4}$

Answer 1

Answer: (B)

Let the statement $P(n)$ be defined as
$P(n)=1.3+2.3^2+3.3^3+\dots ..+n.3^n=\frac{(2n-1)3^{n+1}+3}{4}$
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. $1.3+2.3^2+3.3^3+\dots \dots +k.3^k$
$=\frac{(2k-1)3^{k+1}+3}{4}\dots$(i)
Step III $:$ For $n=k+1$
$\left(1.3+2.3^2+3.3^3+\dots ..+k.3^k\right)+(k+1)3^{k+1}$
$=\frac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k+1}$
[Using equation (i)]
$=\frac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}$
$=\frac{3^{k+1}(2k-1+4k+4)+3}{4}$
[taking $3^{k+1}$ common in first and last term of numerator part $]$
$=\frac{3^{k+1}(6k+3)+3}{4}$
$=\frac{3^{k+1}\cdot 3(2k+1)+3}{4}$
[taking 3 common in first term of numerator part]
$=\frac{3^{(k+1)+1}[2k+2-1]+3}{4}$
$=\frac{[2(k+1)-1]3^{(k+1)+1}+3}{4}$
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 $n$