2020 SJI P1 Q5

The \(r\)th term of a series is given by the expression \(U_r=2^{r+2}-r(r-2)\), where \(r\in\mathbb Z^+\).

Given that \[\sum_{r=1}^n r^2=\frac n6(n+1)(2n+1),\] show that \[\sum_{r=1}^n U_r=8(2^n-1)-\frac n6(n+1)(2n-5).\]

Hence or otherwise, find \[\sum_{r=1}^n\left(2^r-\left(\frac r2\right)\left(\frac r2-1\right)\right)\] giving your answer in terms of \(n\).[8]