Answer: 3, 10, 21, 36

\(u_r = 4r - 1\)

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

First calculate the partial sums from \(S_n=2n^2+n\).

\(S_1=2(1)^2+1=3\)

\(S_2=2(2)^2+2=10\)

\(S_3=2(3)^2+3=21\)

\(S_4=2(4)^2+4=36\)

So the first four sums are \(3,10,21,36\).

The terms of the sequence are found from differences of consecutive sums:

\(u_r=S_r-S_{r-1}\).

Now \(S_r=2r^2+r\), and

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

Hence

\(u_r=(2r^2+r)-\bigl(2(r-1)^2+(r-1)\bigr)\)

\(=2r^2+r-2(r^2-2r+1)-r+1\)

\(=2r^2+r-2r^2+4r-2-r+1\)

\(=4r-1\).

Therefore \(u_r=4r-1\).

For the required sum,

\(\sum_{r=n+1}^{2n} u_r=\sum_{r=n+1}^{2n}(4r-1)\).

There are \(n\) terms, and they form an arithmetic progression with first term \(u_{n+1}=4(n+1)-1=4n+3\) and last term \(u_{2n}=8n-1\).

So

\(\sum_{r=n+1}^{2n} u_r=\frac{n}{2}\bigl((4n+3)+(8n-1)\bigr)\)

\(=\frac{n}{2}(12n+2)=6n^2+n\).