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

Question:

Using the principle of mathematical induction, prove each of the following for all n ϵ N:

$3 \cdot 2^{2}+3^{2} \cdot 2^{3}+3^{3} \cdot 2^{4}+\ldots .+3^{n} \cdot 2^{n+1}=\frac{12}{5}\left(6^{n}-1\right)$

Solution:

To Prove:

$3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots \times+3^{n} \times 2^{n+1}=\frac{12}{5}\left(6^{n}-1\right)$

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

Let $\mathrm{P}(\mathrm{n}): 3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots \times+3^{n} \times 2^{n+1}$

For n = 1

$\mathrm{LHS}=3 \times 2^{2}=12$

$\mathrm{RHS}=\left(\frac{12}{5}\right) \times\left(6^{1}-1\right)$

$=\frac{12}{5} \times 5=12$

Hence, LHS = RHS

P(n) is true for n = 1

Assume P(k) is true

$3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots \times+3^{k} \times 2^{k+1}=\frac{12}{5}\left(6^{k}-1\right)$ ……(1)

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

$3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots \times+3^{k+1} \times 2^{k+2}=\frac{12}{5}\left(6^{k+1}-1\right)$

$3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots \times+3^{k+1} \times 2^{k+2}=\frac{12}{5}\left(6^{k+1}\right)-\frac{12}{5}$

$3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots \times+3^{k} \times 2^{k+1}+3^{k+1} \times 2^{k+2}=$

$\frac{12}{5}\left(6^{k+1}\right)-\frac{12}{5}$

…(2)

We have to prove P(k + 1) from P(k) ie (2) from (1)

From (1)

$3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots \times+3^{k} \times 2^{k+1}=\frac{12}{5}\left(6^{k}-1\right)$

Adding $3^{k+1} \times 2^{k+2}$ both sides

$3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots+3^{k} \times 2^{k+1}+3^{k+1} \times 2^{k+2}$

$=\frac{12}{5}\left(6^{k}-1\right)+3^{k+1} \times 2^{k+2}$

$=\frac{12}{5}\left(6^{k}-1\right)+3^{k} \times 2^{k} \times 12$

$=\frac{12}{5}\left(6^{k}-1\right)+6^{k} \times 12$

$=\left(6^{k}\left(\frac{12}{5}+12\right)-\frac{12}{5}\right)$

$=\left(\frac{72}{5}\right)-\frac{12}{5}$

$=\frac{12}{5}\left(6^{k+1}\right)-\frac{12}{5}$

$3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots+3^{k} \times 2^{k+1}+3^{k+1} \times 2^{k+2}$

$=\frac{12}{5}\left(6^{k+1}\right)-\frac{12}{5}$

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 $x$

Where $n$ is a natural number

Put $k=n-1$

$3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots \times+3^{n} \times 2^{n+1}=\frac{12}{5}\left(6^{n}\right)-\frac{12}{5}$

$3 \times 2^{2}+3^{2} \times 2^{3}+3^{3} \times 2^{4}+\ldots \ldots \times+3^{n} \times 2^{n+1}=\frac{12}{5}\left(6^{n}-1\right)$

Hence proved