Chapter 12.4, Problem 14AYU

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

$1^3\text{ + }{2}^3\text{ + }{3}^3\text{ + }\mathrm{...}\text{ + }{n}^3\text{ = }\frac{1}{4}{n}^2{\left(n\text{ + 1}\right)}^2$

To determine

To prove: The given statement $1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{1}{4} n^2 {\left(n + 1\right)}^2$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 14AYU

As the statement is true for the natural number $\left(k + 1\right)$ terms, hence the statement is true for all natural numbers.

Explanation of Solution

Given:

Statements says the series

$1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{1}{4} n^2 {\left(n + 1\right)}^2$ is true for all natural number.

Formula used:

The Principle of Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: The statement is true for the natural number 1.

CONDITION II: If the statement is true for some natural number $k$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement $1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{1}{4} n^2 {\left(n + 1\right)}^2$   -----(1)

Step 1: Show that statement (1) is true for $n = 1$.

That is $1^3 = \frac{1}{4} 1^2 {\left(1 + 1\right)}^2.$ Hence the statement is true for natural number $n = 1$.

Step 2: Assume that the statement is true for some natural number $k$.

That is $1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{1}{4} n^2 {\left(n + 1\right)}^2$   -----(1)

Step 3: Prove that the statement is true for the next natural number $k + 1$.

That is, to prove that

$\begin{array}{l}{1}^3 + 2^3 + 3^3 + \cdots + {\left(k + 1\right)}^3 = {\frac{1}{4}\left(k + 1\right)}^2 {\left(k + 1 + 1\right)}^2\\ 1^3 + 2^3 + 3^3 + \cdots + k^3 + {\left(k + 1\right)}^3 = \frac{1}{4} {\left(k + 1\right)}^2 {\left(k + 2\right)}^2\end{array}$

Consider $\text{LHS}$ $1^3 + 2^3 + 3^3 + \cdots + k^3 + {\left(k + 1\right)}^3$

$= \frac{1}{4} k^2 {\left(k + 1\right)}^2 + {\left(k + 1\right)}^3 = \frac{1}{4} \left\{{\left(k + 1\right)}^2 \left[k^2 + 4k + 4\right]\right\} = \frac{1}{4} {\left(k + 1\right)}^2 {\left(k + 2\right)}^2$

$\text{LHS} = \text{RHS}$

As the statement is true for the natural number $\left(k + 1\right)$ terms, hence the statement is true for all natural numbers.