2023 DHS Promo Q7
(a)
It is given that $\sum\limits_{r=1}^{n}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)}$.
(i) Find $\sum\limits_{r=1}^{n}{\left( {{2}^{r+1}}+3r-{{r}^{2}} \right)}$ in the form $A\left( {{2}^{n}}-1 \right)+\text{f}\left( n \right)$, where $A$ is a constant and $\text{f}\left( n \right)$ is in fully factorized form.
[3]
(ii) Using your answer in part a(i), find $\sum\limits_{r=4}^{N}{\left( {{2}^{r}}+3r-{{\left( r-1 \right)}^{2}} \right)}$, leaving your answer in the form $B\left( {{2}^{N}} \right)+C\left[ N\left( N-1 \right)\left( 5-N \right) \right]+DN+E$ where $B$, $C$, $D$ and $E$ are constants to be determined.
[4]
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