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

Using the method of differences, sum both sides from \(r=1\) to \(n\):

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

This simplifies to:

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

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

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

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

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

Solving for \(a\):

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

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