`1^2+2^2+3^2++n^2=(n(n+1)(2n+1))/6`

Explanation

Let `P(n):1^2_2^2+3^2+....+n^2=((n+1)(2n+1))/(6)`
Step I For n=1 ,
LHS of Eq. (i) `=1^2=1`
RHS of Eq. (i) `=((1+1)(2xx1+1))/(6)`
`=(1.2.3)/(6)=1`
LHS = RHS
Therefore , P(1) is true .
Step II Let us assume that the result is true for `n=k`. Then , `P(k):1^2+2^2+3^2+......+k^2=(k(k+1)(2k+1))/(6)`
Step III For `n=k+1`, we have to prove that
`P(k+1):1^2+2^2+3^2+......+k^2+(k+1)^2`
`=((k+1)(k+2)(k+3))/(6)`
LHS =`1^2+2^2+3^2+....+k^2+(k+1)^2`
`=(k(k+1)(2k+1))/(6)+(k+1)^2`
`=(k(k+1)(2k+1))/(6)+(k+1)^2`
`=(k+1){(k(2k+1))/(6)+(k+1)}`
`=(k+1){(2k^2+7k+6)/(6)}`
`=(k+1){((k+2)(2k+3))/(6)}=((k+1)(k+2)(2k+3))/(6)= RHS `
This shows that the result is true for `n=k+1`. Therefore , by the principle of methematical induction ,the result is true for all `n in N`.