(a) Expand the expression \(\sum_{r=1}^{N} r(r+1)(3r+4)\) as follows:

\(3 \sum_{r=1}^{N} r^3 + 7 \sum_{r=1}^{N} r^2 + 4 \sum_{r=1}^{N} r.\)

Using standard formulae:

\(\frac{3}{4} N^2 (N+1)^2 + \frac{7}{6} N(N+1)(2N+1) + 2N(N+1).\)

Simplify to:

\(\frac{1}{12} N (9N^3 + 46N^2 + 75N + 38) = \frac{1}{12} N(N+1)(N+2)(9N+19).\)

(b) Express \(\frac{3r+4}{r(r+1)}\) in partial fractions:

\(\frac{3r+4}{r(r+1)} = \frac{4}{r} - \frac{1}{r+1}.\)

Then,

\(\sum_{r=1}^{N} \frac{3r+4}{r(r+1)} \left( \frac{1}{4} \right)^{r+1} = \left( \frac{1}{4^2} \right) \left( 4 - 1 \right) + \left( \frac{1}{4^3} \right) \left( 4 - \frac{1}{2} \right) + \ldots + \left( \frac{1}{4^{N+1}} \right) \left( 4 - \frac{1}{N+1} \right).\)

This simplifies to:

\(\frac{1}{4} - \frac{1}{4^{N+1}(N+1)}.\)

(c) As \(N \to \infty\), the term \(\frac{1}{4^{N+1}(N+1)} \to 0\), so the sum converges to:

\(\frac{1}{4}\).