Recover terms from a cumulative sum · source reference

Printed question

The sum, $S_n$, of the first $n$ terms of a sequence $U_1,U_2,U_3,\ldots$ is given by

$$S_n=\frac n2(c-7n),$$

where $c$ is a constant.

(i) Find $U_n$ in terms of $c$ and $n$. [2]

(ii) Find a recurrence relation of the form $U_{n+1}=f(U_n)$. [2]

Printed working

(i)

$$\begin{aligned}S_n&=\frac n2(c-7n),\\U_n&=S_n-S_{n-1}\\&=\frac n2(c-7n)-\frac{n-1}{2}[c-7(n-1)]\\&=\frac{7+c-14n}{2}.\end{aligned}$$

(ii)

$$U_{n+1}=\frac{7+c-14(n+1)}2=\frac{7+c-14n}2-\frac{14}2=U_n-7.$$

Printed answer key

(i) $\frac{7+c-14n}2$; (ii) $U_n-7$

Mathematical review and qualifications

(i) With $S_0=0$, $U_n=\frac12[n(c-7n)-(n-1)(c-7n+7)]=\frac{c+7-14n}{2}$. This also gives $U_1=(c-7)/2=S_1$.

(ii) Increasing $n$ by one decreases this expression by 7, so $U_{n+1}=U_n-7$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2011/ACJC/Promo/13b] Question PDF pages 1; solution PDF pages 1. The original files remain private QA records outside this native content.