Question

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

Answer 1

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

Step $\left(2\right)$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):1\cdot 2=(1-1)2^{1+1}+2$
$\Rightarrow 2=2$
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.,
$P\left(K\right):1\cdot 2+2\cdot 2^2+3\cdot 2^3+\cdots +K\cdot 2^K$
$=(K-1)2^{K+1}+2\cdots \left(1\right)$

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
$1\cdot 2+2\cdot 2^2+3\cdot 2^3+\cdots +K\cdot 2^K+(K+1)\cdot 2^{K+1}$
$=(K-1)\cdot 2^{K+1}+2+(K+1)\cdot 2^{K+1}$ (Using $\left(1\right)$)
$=2^{K+1}\{K-1+K+1\}+2$
$=2K\cdot 2^{K+1}+2$
$=K\cdot 2^{K+2}+2$
It can be written as
$=\left\{\right(K+1)-1\}\cdot 2^{(K+1)+1}+2$
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 $n\in N$.