Using the principle of mathematical induction, prove each of the following

Solution:

To Prove:

$1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+\ldots \ldots+n \times 2^{n}=(n-1) 2 n+1+2$

Let us prove this question by principle of mathematical induction (PMI)

Let $\mathrm{P}(\mathrm{n}): 1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+\ldots \ldots+n \times 2^{n}$

For $n=1$

LHS $=1 \times 2=2$

RHS $=(1-1) \times 2^{(1+1)}+2$

$=0+2=2$

Hence, LHS $=$ RHS

$P(n)$ is true for $n 1$

Assume $P(k)$ is true

$1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+k \times 2^{k}=(k-1) \times 2^{k+1}+2$ .............(1)

We will prove that P(k + 1) is true

1×

$2^{1}+2 \times 2^{2}+3 \times 2^{3}+(k+1) \times 2^{k+1}=((k+1)-1) \times 2^{(k+1)+1}+2$

$1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+(k+1) \times 2^{k+1}=(k) \times 2^{k+2}+2$

$1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+k 2^{k}+(k+1) \times 2^{k+1}=(k) \times 2^{k+2}+2$ ……(2)

We have to prove P(k + 1) from P(k), i.e. (2) from (1)

From (1)

$1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+k \times 2^{k}=(k-1) \times 2^{k+1}+2$

Adding $(k+1) \times 2^{k+1}$ both sides,

$(1 x$

$\left.2^{1}+2 \times 2^{2}+3 \times 2^{3}+k \times 2^{k}\right)+(k+1) \times 2^{k+1}=(k-1) \times 2^{k+1}+2+$

$(k+1) \times 2^{k+1}$

$=k \times 2^{k+1}-2^{k+1}+2+k \times 2^{k+1}+2^{k+1}$

$=2 k \times 2^{k+1}+2$

$=k \times 2^{k+2}+2$

$\left(1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}+k \times 2^{k}\right)+(k+1) \times 2^{k+1}=k \times 2^{k+2}+2$

Which is the same as $P(k+1)$

Therefore, $P(k+1)$ is true whenever $P(k)$ is true

By the principle of mathematical induction, $P(n)$ is true for

Where $\mathrm{n}$ is a natural number

Put k = n - 1

$\left(1 \times 2^{1}+2 \times 2^{2}+3 \times 2^{3}\right)+n \times 2^{n}=(n-1) \times 2^{n+1}+2$

Hence proved.