A sequence is such that ${{u}_{1}}=p$, where $p$ is a constant, and \({{u}_{n+1}}=\fra

Question

A sequence is such that ${{u}_{1}}=p$, where $p$ is a constant, and ${{u}_{n+1}}=\frac{4}{{{u}_{n}}}$, for $n>0$. Describe how the sequence behaves when $p=2$,[1] $p=3$.[1] Another sequence ${{v}_{1}}$, ${{v}_{2}}$,${{v}_{3}}$ , … is such that ${{v}_{n}}={{v}_{n-1}}+n$, where $n\ge 2$, and ${{v}_{1}}=A$. Find, in terms of $A$ and $n$, an expression for ${{v}_{n}}$,[4] $\sum\limits_{r=1}^{n}{{{v}_{r}}}$. (You need not simplify your answer. You may use the result $\sum\limits_{r=1}^{n}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)}$.)[3] (You need not simplify your answer. You may use the result $\sum\limits_{r=1}^{n}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)}$.)

Accepted Answer

(a)(i) Constant at $2$ (a)(ii) Alternates between $3$ and $\frac43$ (b)(i) $v_n=A+\frac{(n-1)(n+2)}2$ (b)(ii) $\sum_{r=1}^n v_r=n(A-1)+\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}4$

2022 RI P1 Q6

2022 RI P1 Q6

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