A sequence is such that $u_1 = a$ and $u_{n+1} = bu_n$, for $n \geq 1$. Both \(

Question

A sequence is such that $u_1 = a$ and $u_{n+1} = bu_n$, for $n \geq 1$. Both $a$ and $b$ are constants. Explain why the sequence is a geometric progression.[1] Given that $u_2 = 6$, $u_1 + u_2 + u_3 = 26$, and the sum to infinity exists, find the values of $a$ and $b$.[4] It is given that $\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1).$ Find $\sum_{r=1}^{n} r(r+2)$ in terms of $n$. Leave your answer in factorised form.[2] Hence find $\sum_{r=2}^{n} (r+1)(r+3)$ in terms of $n$. You do not need to leave your answer in factorised form.[3]

Accepted Answer

(a)(i) Geometric progression with common ratio $b$ (a)(ii) $a=18$, $b=\dfrac13$ (b)(i) $\dfrac16n(n+1)(2n+7)$ (b)(ii) $\dfrac16(n+1)(n+2)(2n+9)-11$

2025 YIJC P2 Q4

2025 YIJC P2 Q4

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