Question

The sum of the square of the first $n$ natural numbers is $?$

Answer 1

Hint : We find the sum of the square of the first $n$ natural numbers by using the principle of mathematical induction on natural numbers $n$ . First, we prove the desired result for $n = 1$ . Then, we assume the result is true for $n = k$ . Using that we prove the result for $n = k + 1$ . Hence by mathematical induction the result is proved.

Complete step by step solution:
For any natural number $n$, we have to prove that
${1^2} + {2^2} + {3^2} + - - - + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}$ .
Let $p(n)$ be the function of $n$ such that
$p(n) = {1^2} + {2^2} + {3^2} + - - - + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}$ ---(1)
For $n = 1$ in the equation (1), $p(1) = 1 = \dfrac{{1(1 + 1)(2 + 1)}}{6} = 1$ which is true.
Assume that $p(k)$ is true for some positive integer $k$, i.e.,
$p(k) = {1^2} + {2^2} + {3^2} + - - - + {k^2} = \dfrac{{k(k + 1)(2k + 1)}}{6}$ ----(2)
We shall now prove that $p(k + 1)$ is also true. Now, we have
${1^2} + {2^2} + {3^2} + - - - + {k^2} + {(k + 1)^2}$
$= \dfrac{{k(k + 1)(2k + 1)}}{6} + {(k + 1)^2}$
$= \dfrac{{(k + 1)(2{k^2} + k + 6k + 6)}}{6}$
$= \dfrac{{(k + 1)(k + 1 + 1)(2(k + 1) + 1)}}{6}$
Thus $p(k + 1)$ is true, whenever $p(k)$ is true.
Hence, from the principle of mathematical induction, the equation (1) is true for all the natural numbers $n$ .

Note : The principle of mathematical induction is one such tool which can be used to prove a wide variety of mathematical statements. Each such statement is assumed as $p(n)$ associated with positive integers $n$, then proving the statement same as the above procedure.
Also note that we use the mathematical induction only when certain results or statements are formulated in terms of $n$, where $n$ must be a positive integer.
Sum of the first natural numbers $n$ is given by
$1 + 2 + 3 + - - - + n = \dfrac{{n(n + 1)}}{2}$ .