Question

In Problems 1–22, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers $n.$

$1+3+3^2+...+3^{n-1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(3^n-1\right)$

Short answer

The statement$1+3+3^2+...+3^{n-1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(3^n-1\right)$is true for all natural numbers.

Step-by-step solution

Step 1. Given information

The given statement is $1+3+3^2+...+3^{n-1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(3^n-1\right).$We need to prove that the given statement is true for all natural numbers.

Step 2. Proof

Let's prove that the statement is true for $n=1.$

$3^{1-1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(3^1-1\right).$

$3^0=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(2\right).$

$1=1.$

The statement is true for $n=1.$

Let's assume that the statement is true for $n=k.$

$1+3+3^2+...+3^{k-1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(3^k-1\right).$

Let's prove that the statement is true for $n=k+1.$

$1+3+3^2+...+3^{k-1}+3^{k+1-1}.$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(3^k-1\right)+3^k.$

$=\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)·3k-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right..$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(3^{k+1}\right)-1.$

The statement is true for $n=k+1.$

Since the given statement is true for $n=1$and if the statement is true for some $n=k$and it is also true for the next natural number$n=k+1.$

So, the statement is true for all natural numbers.