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

Question

A sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … is such that ${{u}_{n}}=2{{u}_{n-1}}-15{{n}^{2}}+60n+A$, where $A$ is a constant, and $n\ge 2$. It is given that ${{u}_{1}}=2$ and ${{u}_{2}}=4$. Find $A$ and ${{u}_{3}}$.[2] It is known that the $n$th term of this sequence is given by ${{u}_{n}}=p\left( {{2}^{n+1}} \right)-q{{n}^{2}}+r$. Find $p$, $q$ and $r$.[3]

Accepted Answer

(i) $A=-60$, $u_3=-7$ (ii) $p=-\frac{43}{4}$, $q=-15$, $r=30$

2023 JPJC P1 Q1

2023 JPJC P1 Q1

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