(a) Use the standard summation formulas:

\(\sum_{r=1}^{n} 1 = n\)

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

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

Substitute these into the expression:

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

Simplify to get:

\(\frac{1}{3}n - n^2 - \frac{1}{3}n^3\)

(b) Start with the given expression:

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

Verify by expanding and simplifying both sides to show equality.

Use the method of differences:

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

This simplifies to a telescoping series:

\(= \frac{2}{5} - \frac{1}{2} + \frac{3}{10} - \frac{2}{5} + \frac{4}{17} - \frac{3}{10} + \ldots + \frac{n+1}{(n+1)^2 + 1} - \frac{n}{n^2 + 1}\)

Resulting in:

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