Answer: (i) \(\sum_{r=n+1}^{2n} u_r = 14n^3 + 9n^2\).
(ii) \(u_r = 6r^2 - 1\).
Let \(S_n = \sum_{r=1}^{n} u_r\). We are given that \(S_n = n^2(2n+3)\).
(i) The sum from \(r=n+1\) to \(r=2n\) is
\(\sum_{r=n+1}^{2n} u_r = S_{2n} - S_n\).
Now
\(S_{2n} = (2n)^2(2(2n)+3) = 4n^2(4n+3) = 16n^3 + 12n^2\).
So
\(S_{2n} - S_n = (16n^3 + 12n^2) - (2n^3 + 3n^2) = 14n^3 + 9n^2\).
Hence \(\sum_{r=n+1}^{2n} u_r = 14n^3 + 9n^2\).
(ii) Since \(u_r\) is the difference between consecutive partial sums,
\(u_r = S_r - S_{r-1}\).
Now
\(S_r = r^2(2r+3) = 2r^3 + 3r^2\),
and
\(S_{r-1} = (r-1)^2\bigl(2(r-1)+3\bigr) = (r-1)^2(2r+1)\).
So
\(u_r = r^2(2r+3) - (r-1)^2(2r+1)\).
Expanding,
\(r^2(2r+3) = 2r^3 + 3r^2\),
\((r-1)^2(2r+1) = (r^2 - 2r + 1)(2r+1) = 2r^3 - 3r^2 + 1\).
Therefore
\(u_r = (2r^3 + 3r^2) - (2r^3 - 3r^2 + 1) = 6r^2 - 1\).
