It is given that $\sum\limits_{r=1}^{n}{\frac{1}{4{{r}^{2}}-1}}=\frac{n}{2n+1}$. Explain why

Question

It is given that $\sum\limits_{r=1}^{n}{\frac{1}{4{{r}^{2}}-1}}=\frac{n}{2n+1}$. Explain why the series $\sum\limits_{r=1}^{\infty }{\frac{1}{4{{r}^{2}}-1}}$ converges and write down the value to which it converges.[2] Find $\sum\limits_{r=6}^{N}{\frac{1}{\left( 2r+1 \right)\left( 2r+3 \right)}}$ in terms of $N$, express your answer in a single fraction.[3] The sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, $\ldots $ is defined by ${{u}_{1}}=2$, ${{u}_{n+1}}=1-\frac{1}{{{u}_{n}}}$, $n\ge 1$. Find the value of ${{u}_{2}}$, ${{u}_{3}}$ and ${{u}_{4}}$. Hence find the value of $\sum\limits_{r=1}^{50}{{{u}_{r}}}$.[4]

Accepted Answer

(a)(i) $\frac{1}{2}$ (a)(ii) ${\frac{N-5}{13(2N+3)}}$ (b) $\frac{53}{2}$

2024 ACJC Promo Q8

2024 ACJC Promo Q8

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