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prove by mathematical induction 1^2+2^2+3^2.........+n^2= ?[n(n+1)(2n+1)/6]

Precalculus
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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
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