Chapter 12.4, Problem 13AYU

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

$1^2\text{ + }{2}^2\text{ + }{3}^2\text{ + }\mathrm{...}\text{ + }{n}^2\text{ = }\frac{1}{6}n\left(n\text{ + 1}\right)\left(2n\text{ + 1}\right)$

To determine

To prove: The given statement $1^2 + 2^2 + 3^2 + \cdot \cdot \cdot + n^2 = \frac{1}{6}n(n + 1)(2n + 1)$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 13AYU

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

Explanation of Solution

Given:

Statements says the series$1^2 + 2^2 + 3^2 + \cdot \cdot \cdot + n^2 = \frac{1}{6}n(n + 1)(2n + 1)$ 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 $1^2 + 2^2 + 3^2 + \cdot \cdot \cdot + n^2 = \frac{1}{6}n(n + 1)(2n + 1)$   -----(1)

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

That is $1^2 = \frac{1}{6}1(1 + 1)(2.1 + 1).$ 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 $1^2 + 2^2 + 3^2 + \cdot \cdot \cdot + k^2 = \frac{1}{6}k(k + 1)(2k + 1)$   -----(1)

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

That is, to prove that

$1^2 + 2^2 + 3^2 + \cdot \cdot \cdot + {(k + 1)}^2 = \frac{1}{6}(k + 1)(k + 1 + 1)(2(k + 1) + 1)$

$1^2 + 2^2 + 3^2 + \cdot \cdot \cdot + k^2 + {(k + 1)}^2 = \frac{1}{6}(k + 1)(k + 2)(2k + 3)$

Consider LHS $1^2 + 2^2 + 3^2 + \cdot \cdot \cdot + k^2 + {(k + 1)}^2$

$= \frac{1}{6}k(k + 1)(2k + 1) + {(k + 1)}^2 = \frac{1}{6}[(k + 1)[2k^2 + 7k + 6])$

$= \frac{1}{6}(k + 1)(k + 2)(2k + 3)$

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

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