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
 \[1 + 2 + 3 + - - - + n = \dfrac{{n(n + 1)}}{2}\] .