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

Short answer

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

Step-by-step solution

Step 1. Given information

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

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

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

$1=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\times 4.$

$1=1.$

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

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

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

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

$1^3+2^3+3^3+...+k^3+{\left(k+1\right)}^3.$

role="math" localid="1648113249613" $=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.k^2{\left(k+1\right)}^2+{\left(k+1\right)}^3.$

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

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

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.{\left(k+1\right)}^2\left[k^2+2·k+{\left(1\right)}^2\right].$

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

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

So, 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 by the principle of mathematical induction, the statement is true for all natural numbers.