(a) Base case: For \(n = 1\),

\(5 \times 1^4 + 1^2 = \frac{1}{2} (2)^2 (2+1) = 6\), so \(H_1\) is true.

Inductive step: Assume \(\sum_{r=1}^{k} (5r^4 + r^2) = \frac{1}{2} k^2 (k+1)^2 (2k+1)\).

Consider \(\sum_{r=1}^{k+1} (5r^4 + r^2) = \frac{1}{2} k^2 (k+1)^2 (2k+1) + 5(k+1)^4 + (k+1)^2\).

\(= \frac{1}{2} (k+1)^2 (2k^3 + k^2 + 10(k+1)^2 + 2)\)

\(= \frac{1}{2} (k+1)^2 (k^2 + 2k + 1)(2k+3)\)

So \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

(b) Using the result from part (a):

\(5 \sum_{r=1}^{n} r^4 + \frac{1}{6} n(n+1)(2n+1) = \frac{1}{2} n^2 (n+1)^2 (2n+1)\)

\(5 \sum_{r=1}^{n} r^4 = \frac{1}{6} n(n+1)(2n+1)(3n(n+1)-1)\)

\(\sum_{r=1}^{n} r^4 = \frac{1}{30} n(n+1)(2n+1)(3n+3n-1)\)