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

To Prove:

$1+3^{1}+3^{2}+\ldots+3^{n-1}=\frac{3^{n}-1}{2}$

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 $\mathrm{P}(\mathrm{n}): 1+3^{1}+3^{2}+\ldots+3^{n-1}=\frac{3^{n}-1}{2}$

Step 1:

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

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

$\mathrm{P}(\mathrm{k}): 1+3^{1}+3^{2}+\ldots+3^{k-1}=\frac{3^{k}-1}{2}$

Now,

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

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

$={ }^{(k)}\left(\frac{1}{2}+1\right)-\frac{1}{2}$

$=3^{(k)}\left(\frac{3}{2}\right)-\frac{1}{2}$

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

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

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

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

Hence, by the principle of mathematical induction, we have

$1+3^{1}+3^{2}+\ldots+3^{n-1}=\frac{3^{n}-1}{2}$ for all $\mathrm{n} \in \mathrm{N}$