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 $\mathrm{n} \in \mathrm{N}$ :

$2+6+18+\ldots+2 \times 3^{n-1}=\left(3^{n}-1\right)$

Solution:

To Prove:

$2+6+18+\ldots+2^{\times} 3^{n-1}=\left(3^{n}-1\right)$

Steps to prove by mathematical induction:

Let $P(n)$ be a statement involving the natural number $n$ such that

(i) $P(1)$ is true

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

Then $P(n)$ is true for all $n \in N$

Therefore,

Let $\mathrm{P}(\mathrm{n}): 2+6+18+\ldots+2 \times 3^{n-1}=\left(3^{n}-1\right)$

Step 1

$P(1)=3^{1}-1=3-1=2$

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

$P(k): 2+6+18+\ldots+2^{\times} 3^{k-1}=\left(3^{k}-1\right)$

Now,

$2+6+18+\ldots+2 \times 3^{k-1}+2 \times 3^{k+1-1}=\left(3^{k}-1\right)+2 \times 3^{k}$

$=-1+3 \times 3^{\mathrm{k}}$

$=3^{\mathrm{k}+1-1}$

$=\mathrm{P}(\mathrm{k}+1)$

Hence, P(k + 1) is true whenever P(k) is true

Hence, by the principle of mathematical induction, we have

$2+6+18+\ldots+2^{\times} 3^{n-1}=\left(3^{n}-1\right)$ for all $n \in N$