Question

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

Answer 1

Let the given statement be $P\left(n\right)$ i.e.,
$P\left(n\right)=1+3+3^2+......+3^{n-1}=\frac{(3^n-1)}{2}$
For $n=1$, we have
$P\left(1\right)=\frac{(3^1-1)}{2}=\frac{3-1}{2}=\frac{2}{2}=1$, which is true.
Let $P\left(k\right)$ be true for some positive integer $k$, i.e.,
$1+3+3^2+........+3^{k-1}=\frac{(3^k-1)}{2}..........\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Now $P(k+1)=1+3+3^2+........+3^{k-1}+3^{(k-1)+1}$
$=(1+3+3^2+.....+3^{k-1})+3^k$
$=\frac{(3^k-1)}{2}+3^k$ [Using $\left(i\right)$]
$=\frac{(3^k-1)+{2.3}^k}{2}$
$=\frac{3.3^k-1}{2}$
$=\frac{3^{k+1}-1}{2}$
Thus $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$.