A sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … is such that \({{u}_{n+2}}=A{{u}_{n+1

Question

A sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … is such that ${{u}_{n+2}}=A{{u}_{n+1}}-{{u}_{n}}$, where $A$ is a positive constant and $n\ge 1$. Given that ${{u}_{1}}=1$, ${{u}_{3}}=5$ and ${{u}_{4}}=13$, find $A$ and ${{u}_{2}}$.[4] The sum of the first $n$ terms of a sequence ${{v}_{1}}$, ${{v}_{2}}$, ${{v}_{3}}$, … is given by $2{{n}^{2}}+4n$. Find an expression for ${{v}_{n}}$, in terms of $n$, in simplified form.[2] Hence find ${{v}_{2}}+{{v}_{4}}+{{v}_{6}}+...+{{v}_{48}}$.[3]

Accepted Answer

(a) A = 3, u₂ = 2 (b)(i) vₙ = 4n + 2 (b)(ii) 2448

2022 VJC J1 MYE Q5

2022 VJC J1 MYE Q5

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