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

Answer
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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
 12+22+32++n2=n(n+1)(2n+1)6 .
Let p(n) be the function of n such that
 p(n)=12+22+32++n2=n(n+1)(2n+1)6 ---(1)
For n=1 in the equation (1), p(1)=1=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)=12+22+32++k2=k(k+1)(2k+1)6 ----(2)
We shall now prove that p(k+1) is also true. Now, we have
 12+22+32++k2+(k+1)2
 =k(k+1)(2k+1)6+(k+1)2
 =(k+1)(2k2+k+6k+6)6
 =(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=n(n+1)2 .
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