Prove by the principle of mathematical induction that for all `n in N :` `1^2+2^2+3^2++n^2=1/6n(n+1)(2n+1)`

Explanation

Let P (n) :
`1^(2)+2^(2)+3^(3)+…….+n^(2)=1/6 n(n +1)(2n+1)`
For n=1
`L.H.S. = 1^(2)=1`
`R.H.S. =1/6 .1. (1+1) .(2.2+1)`
`=1/6 .2.3=1`
`:. L.H.S. =R.H.S.`
Therefore P (n) is true for n=1
Let p (n) is true for n=K.
`P(K) :1^(2)+2^(2)+3^(3)+.......+K^(2)`
`=1/6 k (K +1) (2K+1)`
Adding `(K+1)^(2)` on both sides
`1^(2)+2^(2)+3^(2)+........+K^(2)+(K+1)^(2)`
`=1/6 K(K+1)(2K+1)+(k+1)^(2)`
`=(K(K+1)(2K+1)+6(K+1)^(2))/(6)`
`=1/6 (K+1)[K(2K+1)+6(K+1)]`
`=1/6 (K+1)(2K^(2)+7K+6)`
`=1/6 (K+1)(K+2)(2K+3)`
`=1/6 (K+1){(K+1)+1}{2(K+1)+1}`
`rArr " "P (n) "is also true for" n =K =1`
Hence by the principle of mathematical induction given statement is true for all natural numbers 'n'