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 $\sum\limits_{r=1}^{n}{6r\left( r+1 \right)}$, in terms of $n$.[2] Express $\frac{1}{r\left( r+1 \right)\left( r+2 \right)}$ in partial fractions.[1] Hence find ${{S}_{N}}$ in terms of $N$, where ${{S}_{N}}=\sum\limits_{r=1}^{N}{\frac{1}{r(r+1)(r+2)}}$, and deduce that ${{S}_{N}}<\frac{1}{4}$.[6]

Accepted Answer

(a) $2n(n+1)(n+2)$. (b)(i) $\frac1{2r}-\frac1{r+1}+\frac1{2(r+2)}$. (b)(ii) Since $\frac{1}{\left( N+1 \right)\left( N+2 \right)}>0$, for $N\in {{\mathbb{Z}}^{+}}$, ${{S}_{N}}<\frac{1}{4}$.

2023 SAJC Promo Q3

2023 SAJC Promo Q3

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