Question

Prove that \[{1^2} + {2^2} + ..... + {n^2} > \dfrac{{{n^3}}}{3}\].

Answer 1

Hint: Use the principle of mathematical induction to prove \[{1^2} + {2^2} + ..... + {n^2} > \dfrac{{{n^3}}}{3}\]. First, check if it is satisfied for n = 1. Then assume it is true for n = k and then prove that it is true for n = k + 1. Hence, conclude, it is true for all n.

Complete step-by-step answer:
A class of integers is called hereditary if, for any integer, x belongs to the class, then the successor of x also belongs to the class.

The principle of mathematical induction states if the integer belongs to the class and the class is hereditary, then every non-negative integer belongs to the class.

As the first step in proving a statement using mathematical induction, the first term of the statement is checked. In most cases, it is using the integer 1. If it is true, then we proceed or else we conclude that the statement is false.

In the next step, we assume the statement is true for an integer k and prove that it is true for the integer k + 1. If we can prove, then the statement is true for all non-negative integers.

Let \[P(n):{1^2} + {2^2} + ..... + {n^2} > \dfrac{{{n^3}}}{3}\]

We check if P(1) is true.

\[P(1):1 > \dfrac{1}{3}\]

Hence, P(1) is true, then we assume that P(k) is true.

\[P(k):{1^2} + {2^2} + ..... + {k^2} > \dfrac{{{k^3}}}{3}..........(1)\]

Now, we proceed to prove the following statement:

\[P(k):{1^2} + {2^2} + ..... + {k^2} + {(k + 1)^2} > \dfrac{{{{(k + 1)}^3}}}{3}\]

Consider equation (1), we add \[{(k + 1)^2}\] to both sides of the equation (1) to get as follows:

\[{1^2} + {2^2} + ..... + {k^2} + {(k + 1)^2} > \dfrac{{{k^3}}}{3} + {(k + 1)^2}\]

We know that \[{(a + b)^2} = {a^2} + 2ab + {b^2}\], then we have:

\[{1^2} + {2^2} + ..... + {k^2} + {(k + 1)^2} > \dfrac{{{k^3}}}{3} + ({k^2} + 2k + 1)\]

Simplifying the expression, we have:

\[{1^2} + {2^2} + ..... + {k^2} + {(k + 1)^2} > \dfrac{{{k^3} + 3{k^2} + 6k + 3}}{3}\]

We know that \[{(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\], then we have:

\[{1^2} + {2^2} + ..... + {k^2} + {(k + 1)^2} > \dfrac{{{{(k + 1)}^3} + 3k + 2}}{3}\]

Splitting the terms, we get:

\[{1^2} + {2^2} + ..... + {k^2} + {(k + 1)^2} > \dfrac{{{{(k + 1)}^3}}}{3} + \dfrac{{3k + 2}}{3}\]

We know that \[\dfrac{{3k + 2}}{3}\] is positive for a non-negative integer k. Then, we have:

\[{1^2} + {2^2} + ..... + {k^2} + {(k + 1)^2} > \dfrac{{{{(k + 1)}^3}}}{3} + \dfrac{{3k + 2}}{3} >

\dfrac{{{{(k + 1)}^3}}}{3}\]

\[{1^2} + {2^2} + ..... + {k^2} + {(k + 1)^2} > \dfrac{{{{(k + 1)}^3}}}{3}\]

Hence, we proved that P(k+1) is also true.

Hence, the statement is true for all non-negative integers.

Note: You can also prove the statement using the sum of first n natural numbers which is \[{1^2} + {2^2} + ..... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}\] and simplifying it.