(a) Use the standard summation formulas:

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

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

Substitute into the expression:

\(6 \left( \frac{1}{6}n(n+1)(2n+1) \right) + 6 \left( \frac{1}{2}n(n+1) \right) - 5n\)

Simplify to get \(2n^3 + 6n^2 - n\).

(b) Divide by the denominator:

\(\frac{6r^2 + 6r - 5}{r^2 + r} = 6 - \frac{5}{r(r+1)}\)

Find partial fractions:

\(\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}\)

Sum the series:

\(\sum_{r=1}^{n} \left( 1 - \frac{1}{r+1} \right) = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} - \frac{1}{n+1}\)

Result: \(6n - 5 + \frac{5}{n+1}\).

(c) Use the result from (b):

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

\(= 12n - 5 + \frac{5}{2n+1} - (6n - 5 + \frac{5}{n+1})\)

Simplify to get \(6n - \frac{5n}{(n+1)(2n+1)}\).