Derive the formula
l2 + 22 + 32 + ... + n2 = n(n + 1) (2n +1)/6
Using the following steps.
(a) Show that (k + l)3 - k3 = 3k2 + 3k + 1.
(b) Show that
[23 - 13] + [33 - 23] + [43 - 33] + ... + [(n + 1)3 - n3] = (n + 1)3 - 1
(c) Apply (a) to each term on the left side of (b) to show that
(n + l)3 - l = 3[12 + 22 + 32 + ... + n2] + 3[1 + 2 + 3 + ... + n] + n
(d) Solve the equation in (c) for 12 + 22 + 32 + ... + n2, use the result of Exercise 5, and then simplify.