It is given that \(\sum\limits_{r=1}^{n}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \r

Question

It is given that $\sum\limits_{r=1}^{n}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)}$. Find $\sum\limits_{r=1}^{n}{\left( {{2}^{r+1}}+3r-{{r}^{2}} \right)}$ in the form $A\left( {{2}^{n}}-1 \right)+\mathrm{f}\left( n \right)$, where $A$ is a constant and $\mathrm{f}\left( n \right)$ is in fully factorized form.[3] 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]

Accepted Answer

(a)(i) $4(2^n-1)+\dfrac{n(n+1)(4-n)}3$, so $A=4$ and $\mathrm{f}(n)=\dfrac{n(n+1)(4-n)}3$ (a)(ii) $B=2$, $C=\dfrac13$, $D=3$, $E=-29$

2023 DHS Promo Q7

2023 DHS Promo Q7

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