Question

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

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i. e.,
$P\left(n\right):1+3+3^2+\cdots +3^{n-1}=\frac{(3^n-1)}{2}$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):1=\frac{(3^1-1)}{2}$
$\Rightarrow 1=\frac{2}{2}$
$\Rightarrow 1=1$
Thus $P\left(n\right)$ is true for $n=1$.

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$ in $P\left(n\right)$ and assume this is true for some natural number $K$ i.e.,
$P\left(K\right):1+3+3^2+\cdots +3^{K-1}=\frac{(3^K-1)}{2}\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now, we shall prove that $P(K+1)$ is true
whenever $P\left(K\right)$ is true.
Now, we have
$1+3+3^2+\cdots +3^{K-1}+3^K$
$=(1+3+3^2+\cdots +3^{K-1})+3^K$
$=\frac{(3^K-1)}{2}+3^K$ (Using $\left(1\right)$)
$=\frac{3^K-1+2\cdot 3^K}{2}$
$=\frac{3\cdot 3^K-1}{2}$
$=\frac{3^{K+1}-1}{2}$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Final Answer :
Hence, from the principle of Mathematical induction, the statement $P\left(n\right)$ is true for all $n\in N$.