Discrete Math - Mathematical Induction Help

Prove by the principle of mathematical induction.

1.) The sum of the consecutive fourth powers in Z+ is: S4(n) = n(n+1)(2n+1)(3n^2+3n-1) / 30
(***The S4 should be S subscript 4)

2.) The sum of consecutive 5th powers in Z+ is: S5(n) = n^2(n+1)^2(2n^2+2n-1) / 12
(****The S5 should be S subscript 5)

In a word: tedious.

1) This is true for n = 1, because 1^4 = 1 = 1 * 2 * 3 * 5 / 30.

Now, assume that the claim is true for n = k.

We want to show that
S4(k+1) = (1/30) (k+1)((k+1) + 1)(2(k+1) + 1) [3(k+1)^2 + 3(k+1) - 1]
..............= (1/30) (k+1)(k+2)(2k+3)(3k^2 + 9k + 5).

Then S4(k+1)
= (1^4 + 2^4 + ... + k^4) + (k+1)^4
= [k(k+1)(2k+1)(3k^2+3k-1) / 30] + (k+1)^4, by inductive hypothesis
= (1/30)(k+1) * [k(2k+1)(3k^2+3k-1) + 30(k+1)^3]
= (1/30)(k+1) * (6k^4 + 39k^3 + 91k^2 + 89k + 30), by algebra.

Now, we look for factors for the quartic.
By the rational root theorem, the possible rational roots are of the form a/b where a is a factor of the constant term 30 and b is a factor of the leading coefficient 6.

However, from writing out what we need to show, we know that from the factors
(k+2)(2k+3) [having already accounted for k+1], that (k+2) and (2k+3) should be factors.

Indeed this is the case, because repeated division yields
6k^4 + 39k^3 + 91k^2 + 89k + 30 = (k + 2)(6k^3 + 27k^2 + 37k + 15).
and 6k^3 + 27k^2 + 37k + 15 = (2k+3)(3k^2 + 9k + 5).

Hence, S4(k+1) = (1/30) (k+1)(k+2)(2k+3)(3k^2 + 9k + 5), as required.
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Now try to do #2 using the same strategy (write out S5(k+1) so you know what the factors should be in advance.)

I hope this helps!