Prove by induction that 1^2 + 2^2 + 3^2 + . . . + n^2 = (1/6)n(n+1)(2n+1)
Base case - First let us consider the base case where n=1. Then the left hand side of the equation becomes 1 and the right hand side is (1/6)12*3=6/6=1. Thus LHS=RHS and we are done.Induction hypothesis - Now let us assume 1^2 + 2^2 + 3^2 + . . . + k^2 = (1/6)k(k+1)(2k+1) is true for some integer value of k.Induction step - Consider the case where n=k+1. Then LHS becomes 1^2 + 2^2 + 3^2 + . . . + k^2 +(k+1)^2 = (1/6)k(k+1)(2k+1) + (k+1)^2 =(k+1)((1/6)k(2k+1)+(k+1))=(1/6)(k+1)(k(2k+1)+6(k+1))=(1/6)(k+1)(2k^2+k+6k+6)=(1/6)(k+1)(k+2)(2k+3) by induction hypothesis. But this is the equation above with k replaced with (k+1).Thus, if the statement is true for some integer k, it must also be true for k+1, k+2, ..., that is for all integers greater than it.But since it is true for 1, it must be true for all integers by mathematical induction,
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