(a) Consider \((r+1)^2 - r^2 = r^2 + 2r + 1 - r^2 = 2r + 1\).

Using the method of differences, sum both sides:

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

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

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

(b) \(\sum_{r=1}^{n} (r+a) = \sum_{r=1}^{n} r + an\)

\(\frac{1}{2} n(n+1) + an = n\)

Solving for \(a\):

\(a = \frac{1}{2}(1-n)\)