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

Question

Given that $\sum\limits_{r=1}^{n}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)}$, find $\sum\limits_{r=7}^{n+1}{\left( {{2}^{r}}+{{r}^{2}}-r \right)}$ in terms of $n$.[4] Use the method of differences to show that $\sum\limits_{r=2}^{n}{\frac{1}{{{r}^{2}}-1}}=\frac{3}{4}+\frac{A}{n}+\frac{A}{n+1}$, where $A$ is a constant to be determined.[3] Explain why the series $\sum\limits_{r=2}^{\infty }{\frac{1}{{{r}^{2}}-1}}$ converges, and write down its value.[2] Hence deduce that $\frac{2}{{{2}^{2}}}+\frac{2}{{{3}^{2}}}+\frac{2}{{{4}^{2}}}+...$ is less than $\frac{3}{2}$.[2]

Accepted Answer

${{2}^{n+2}}+\frac{1}{3}n\left( n+1 \right)\left( n+2 \right)-198$

2021 ACJC Promo Q9

2021 ACJC Promo Q9

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