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
 $$1^2+2^2+3^2+---+n^2={\displaystyle \frac{n(n+1)(2n+1)}{6}}$$ .
Let $$p(n)$$ be the function of $$n$$ such that
 $$p(n)=1^2+2^2+3^2+---+n^2={\displaystyle \frac{n(n+1)(2n+1)}{6}}$$ ---(1)
For $$n=1$$ in the equation (1), $$p(1)=1={\displaystyle \frac{1(1+1)(2+1)}{6}}=1$$ which is true.
Assume that $$p(k)$$ is true for some positive integer $$k$$, i.e.,
 $$p(k)=1^2+2^2+3^2+---+k^2={\displaystyle \frac{k(k+1)(2k+1)}{6}}$$ ----(2)
We shall now prove that $$p(k+1)$$ is also true. Now, we have
 $$1^2+2^2+3^2+---+k^2+(k+1{)}^2$$
 $$={\displaystyle \frac{k(k+1)(2k+1)}{6}}+(k+1{)}^2$$
 $$={\displaystyle \frac{(k+1)(2k^2+k+6k+6)}{6}}$$
 $$={\displaystyle \frac{(k+1)(k+1+1)(2(k+1)+1)}{6}}$$
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
 $$1+2+3+---+n={\displaystyle \frac{n(n+1)}{2}}$$ .