for n=1,
L.H.S = 1^2 = 1, R.H.S= [1*(1+1)(2*1+1)/6]=1
Hence, L.H.S= R.H.S for n=1
Now assume the result is true for some integer p which implies
\[1^{2}+ 2^{2}+...+p ^{2} = p(p+1)(2p+1)/6 + \]
Now add \[(p+1)^{2}\] for both sides.
\[[1^{2}+ 2^{2}+...+p ^{2} + (p+1)^{2} = p(p+1)(2p+1)/6 + (p+1)^{2}\]
Now you can simplify the R.H.S of the hypothesis to obtain \[(p+1)(p+2)(2(p+1)+1)/6\] which implies that the given expression is valid for \[(p+1)^{th}\] integer given that the result is true for the \[p ^{th}\] integer. Hence the result stands for every positive integer according to mathematical induction