Question

Prove the following by using the principle of mathematical induction for all $n\in N$
$1\cdot 3+2\cdot 3^2+3\cdot 3^3+\cdots +n\cdot 3^n=\frac{(2n-1)3^{n+1}+3}{4}$

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):1\cdot 3+2\cdot 3^2+3\cdot 3^3+\cdots +n\cdot 3^n=\frac{(2n-1)3^{n+1}+3}{4}$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):1.3=\frac{(2\times 1-1)3^{1+1}+3}{4}$
$\Rightarrow 3=\frac{1\cdot 9+3}{4}=\frac{12}{4}=3$
$\Rightarrow 3=3$
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$ in $P\left(n\right)$ and assume this is true for some natural number $K$ i.e.,
$P\left(K\right):1\cdot 3+2\cdot 3^2+3\cdot 3^3+\cdots +K\cdot 3^K$
$=\frac{(2K-1)3^{K+1}+3}{4}\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now, we have
$1\cdot 3+2\cdot 3^2+3\cdot 3^3+\cdots +K\cdot 3^K+(K+1)3^{K+1}$
$=\frac{(2K-1)3^{K+1}+3}{4}+(K+1)3^{K+1}$ (Using $\left(1\right)$)
$=\frac{(2K-1)3^{K+1}+3+4(K+1)3^{K+1}}{4}$
$=\frac{3^{K+1}(2K-1+4K+4)+3}{4}$
$=\frac{3^{K+1}(6K+3)+3}{4}$
$=\frac{3\cdot 3^{K+1}(2K+1)+3}{4}$
$=\frac{(2K+1)3^{K+2}+3}{4}$
We can write it as,
$=\frac{\left(2\right(K+1)-1)3^{\left(\right(K+1)+1)}+3}{4}$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Final Answer :
Hence, from the principle of mathematical induction, the statement $P\left(n\right)$ is true for all $n\in N$.