by way of fact of certainty the effective integers initiate up at n = a million, that's our base case. With n = a million, we see that the LHS is two^2 = 4 on an analogous time by way of fact the RHS is two(a million)(a million + a million)(2 + a million)/3 = 4. This exhibits that the backside case is authentic. The inductive step is to coach that P(n + a million) holds if P(n) holds. it extremely is: 2^2 + 4^2 + 6^2 + ... + (2n)^2 + [2(n + a million)]^2 = 2(n + a million)[(n + a million) + a million][2(n + a million) + a million]/3, given 2^2 + 4^2 + 6^2 + ... + (2n)^2 = 2n(n + a million)(2n + a million)/3. Then, utilising this: 2^2 + 4^2 + 6^2 + ... + (2n)^2 + [2(n + a million)]^2 = [2^2 + 4^2 + 6^2 + ... + (2n)^2] + [2(n + a million)]^2 = 2n(n + a million)(2n + a million)/3 + [2(n + a million)]^2, via the inductive hypothesis = 2n(n + a million)(2n + a million)/3 + 4(n + a million)^2 = [2n(n + a million)(2n + a million) + 12(n + a million)^2]/3 = 2(n + a million)[n(2n + a million) + 6(n + a million)]/3 = 2(n + a million)(2n^2 + 7n + 6)/3 = 2(n + a million)(n + 2)(2n + 3)/3, via factoring = 2(n + a million)[(n + a million) + a million][2(n + a million) + a million]/3. by way of fact of certainty the backside case and the inductive step carry, we are able to end the info. i desire this helps!