Recurrence relations: EJC 2024 Promo Q11

(ai)

$$\begin{aligned} u_2 &= \frac{1}{5}u_1-3=\frac{1}{5}(2)-3=-2.6\quad\text{or }-\frac{13}{5}\\ u_3 &= \frac{1}{5}u_2-3=\frac{1}{5}(-2.6)-3=-3.52\quad\text{or }-\frac{88}{25}\\ u_4 &= \frac{1}{5}u_3-3=\frac{1}{5}(-3.52)-3=-3.704\quad\text{or }-\frac{463}{125} \end{aligned}$$

(aii)

Since sequence converges to $l$, $u_n,u_{n+1}\approx l$ when $n$ is large.

Solving $l=\dfrac{l}{5}-3$, we get $l=-3.75$ or $-\dfrac{15}{4}$.

(bi)

$\displaystyle\sum_{r=1}^{n}u_{r+4}$ is the sum of $n$ terms of an arithmetic progression with first term $u_5=2-3(4)=-10$ and common difference $d=-3$. $$\begin{aligned} \therefore\ \sum_{r=1}^{n} u_{r+4} &= \frac{n}{2}\big[2(-10)+(n-1)(-3)\big]\\ &= -\frac{n(3n + 17)}{2} \end{aligned}$$

(bii)

Since $u_n = 2+(n-1)(-3)=5-3n$, $$\begin{aligned} u_{3n+4}&=5-3(3n+4)=-7-9n\\ u_{3(n+1)+4}=u_{3n+7}&=5-3(3n+7)=-16-9n\\ \text{So }u_{3n+7}-u_{3n+4}&=(-16-9n)-(-7-9n)=-9 \end{aligned}$$ This is a constant independent of $n$, so the sequence is an A.P.

(biii)

$\displaystyle\sum_{r=1}^{100} u_{3r+4} = u_7 + u_{10} + \cdots + u_{304}$ is the sum of $100$ terms of an A.P. with first term $u_7 = 2 - 3(6) = -16$ and common difference $-9$. $$\begin{aligned} \therefore\ \sum_{r=1}^{100} u_{3r+4} &= \frac{100}{2}\big[2(-16) + 99(-9)\big]\\ &= -46150. \end{aligned}$$