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+4+4^2+...+4^{n-1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(4^n-1\right)$

Short answer

The statement$1+4+4^2+...+4^{n-1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(4^n-1\right)$is true for all natural numbers.

Step-by-step solution

Step 1. Given information

The given statement is $1+4+4^2+...+4^{n-1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(4^n-1\right).$We need to prove that the given statement is true for all natural numbers by using mathematical induction.

Step 2. Proof

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

$4^{1-1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(4^1-1\right).$

$4^0=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(3\right).$

$1=1.$

So, the statement is true for $n=1.$

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

So, role="math" localid="1648054611339" $1+4+4^2+...+4^{k-1}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(4^k-1\right).$

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

$1+4+4^2+...+4^{k-1}+4^{k+1-1}.$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(4^k-1\right)+4k.$

$=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.·4k-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right..$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(4^{k+1}-1\right).$

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$k+1.$

So, the statement is true for all natural numbers.