A sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … is given by ${{u}_{n}}=\frac{1}{4}\left( 7-{{3}^

Question

A sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … is given by ${{u}_{n}}=\frac{1}{4}\left( 7-{{3}^{n}} \right)$, for $n\ge 1$. By expressing $4{{u}_{n+1}}$ and $12{{u}_{n}}$ in terms of $n$, or otherwise, find a recurrence relation of the form ${{u}_{n+1}}=\mathrm{f}\left( {{u}_{n}} \right)$. [3] Find $\sum\limits_{r=2}^{n}{{{u}_{r}}}$. [3] The sum, ${{S}_{n}}$, of the first $n$ terms of a sequence ${{v}_{1}}$, ${{v}_{2}}$, ${{v}_{3}}$, … is given by ${{S}_{n}}=2-\frac{1}{\left( n+1 \right)!}$. Find a formula for ${{v}_{n}}$. [1] Explain why the series $\sum\limits_{r=1}^{\infty }{{{v}_{r}}}$ converges. Hence, state the value of the sum to infinity. [2]

Accepted Answer

(a)(i) $u_{n+1} = 3u_n - \frac{7}{2}$, (a)(ii) $\frac{7}{4}n - \frac{9}{8}(3^{n-1}) - \frac{5}{8}$, (b)(i) $V_n = \frac{1}{n!} - \frac{1}{(n+1)!}$, (b)(ii) $2$

2024 ACJC J1 WA2 Q3

2024 ACJC J1 WA2 Q3

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