(b) 1/6k(k + 1)(2k + 1) (a) 1/6k(k + 1)(2k + 1) + (k + 1)^2
Expanding b with the substitution, we have
(1 / 6) (k + 1) ( [k + 1] + 1) (2 [k + 1] + 1) =
(1 /6) ( k + 1) (k + 2) (2k + 3) =
(1/6) ( k^2 + 3k + 2)(2k + 3) =
(1 / 6) (2k^3 + 6k^2 + 4k + 3k^2 + 9k + 6) =
(1 /6) (2k^3 + 9k^2 + 13k + 6) (1)
Expanding a, we have
1/6k(k + 1)(2k + 1) + (k + 1)^2 =
(1 / 6) (k^2 + k)(2k + 1) + k^2 + 2k + 1 =
(1/6) ( 2k^3 + 2k^2 + k^2 + k) + k^2 + 2k + 1 =
(1/6) (2k^3 + 2k^2 + k^2 + k) + (1/6)(6k^2 + 12k + 6) =
(1 / 6) ( 2k^3 + 9k^2 + 13k + 6) (2)
And since (1) = (2).....the expressions are equivalent
