Question

Prove the following by 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

Given,

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

Let $P\left(n\right)=1+3+3^2+.......+3^{n-1}=\frac{3^n-1}{2}$

For $n=1$

LHS=1

RHS=

$\frac{3^1-1}{2}=\frac{2}{2}=1$

LHS=RHS

Therefore P(n) is true for $n=1$

Assume that P(k) is true.

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

we have to prove that $P(k+1)$ is true.

add $3^k$ on both sides, we get,

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

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

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

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

Therefore $P(k+1)$ is true when P(k) is true.

Therefore by the principle of mathematical induction, P(n) is true for n.