(a) Base case: For \(n = 1\), \(\sum_{r=1}^{1} r^2 = 1^2 = 1\) and \(\frac{1}{6} \times 1 \times 2 \times 3 = 1\). So \(H_1\) is true.

Inductive step: Assume \(\sum_{r=1}^{k} r^2 = \frac{1}{6}k(k+1)(2k+1)\) is true for \(n = k\).

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

\(= \frac{1}{6}k(k+1)(2k+1) + (k+1)^2\)

\(= \frac{1}{6}(k+1)(2k^2 + k + 6k + 6)\)

\(= \frac{1}{6}(k+1)(2k^2 + 7k + 6)\)

\(= \frac{1}{6}(k+1)(k+2)(2k+3)\)

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

(b) Using the identity, expand and simplify:

\((2n+1)^5 - 1\)

\(= 160S_n + \frac{40}{3}n(n+1)(2n+1) + 2n\)

\(160S_n = (2n+1)^5 - \frac{40}{3}n(n+1)(2n+1) - (2n+1)\)

\(160S_n = (2n+1)((2n+1)^4 - \frac{40}{3}n(n+1) - 1)\)

\(\Rightarrow S_n = \frac{1}{30}n(n+1)(2n+1)(3n^2 + 3n - 1)\)

(c) \(n^{-5}S_n = \frac{1}{30}(1 + \frac{1}{n})(2 + \frac{1}{n})(3 + \frac{3}{n} - \frac{1}{n^2})\)

As \(n \to \infty\), \(n^{-5}S_n \to \frac{1}{5}\).