Answer: \(\sum_{r=n+1}^{2n} r^2 = \frac{1}{6}n(2n+1)(7n+1)\).
Write the required sum as the difference between two standard sums:
\(\sum_{r=n+1}^{2n} r^2 = \sum_{r=1}^{2n} r^2 - \sum_{r=1}^{n} r^2\).
Now use the formula \(\sum_{r=1}^{m} r^2 = \frac{m(m+1)(2m+1)}{6}\).
So
\(\sum_{r=1}^{2n} r^2 = \frac{(2n)(2n+1)(4n+1)}{6}\)
and
\(\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\).
Hence
\(\sum_{r=n+1}^{2n} r^2 = \frac{(2n)(2n+1)(4n+1)}{6} - \frac{n(n+1)(2n+1)}{6}\).
Factor out \(\frac{1}{6}n(2n+1)\):
\(\sum_{r=n+1}^{2n} r^2 = \frac{1}{6}n(2n+1)\bigl(2(4n+1)-(n+1)\bigr)\).
Simplifying the bracket gives
\(2(4n+1)-(n+1)=8n+2-n-1=7n+1\).
Therefore
\(\sum_{r=n+1}^{2n} r^2 = \frac{1}{6}n(2n+1)(7n+1)\).
