Frederick, a social media influencer, is starting a new online account at the start of a month. From

Question

Frederick, a social media influencer, is starting a new online account at the start of a month. From the second month onwards:

His organic followers at the end of each month will be $a$ times the total followers he had at the end of the previous month, where $a$ is a positive constant.
A company he engaged in will provide him with $1000$ additional followers in the middle of each month.

Let $F(n)$ denote the total number of organic and additional followers Frederick had at the end of $n$ months after he started his new online account, for $n \in \mathbb{Z}^+$.

Write down a recurrence relationship between $F(n)$ and $F(n+1)$ for $n \in \mathbb{Z}^+$, giving your answer in terms of $a$.[2]

It is found that Frederick has $3500$ followers at the end of the first month.

Show that $F(3)=3500a^2+1000a+1000$.[2]
Find an expression for $F(n)$. Hence, find the number of months required for Frederick’s followers to exceed one million if $a=1.5$.[4]
Given that the number of followers exceeds $20000$ by the end of the first year, determine the range of values of $a$.[2]

Frederick finds out that, instead of the additional $1000$ followers in the middle of every month, the company can only provide him with additional $b$ followers in the middle of every month.

Given that the number of followers remains constant since the end of the first month, find the relationship between $a$ and $b$.[2]

Accepted Answer

(a) $\mathrm{F}(n+1)=a\mathrm{F}(n)+1000$ (b) $\mathrm{F}(3)=3500a^2+1000a+1000$ (c) $\mathrm{F}(n)=3500a^{n-1}+\dfrac{1000(a^{n-1}-1)}{a-1}$ for $a\ne1$; $\mathrm{F}(n)=3500+1000(n-1)$ for $a=1$; 14 months (d) $a>\alpha$, where $\alpha=1.0485831335\ldots$ solves $3500\alpha^{11}+1000\sum_{j=0}^{10}\alpha^j=20\,000$ (e) $b=3500(1-a)$

2025 HCI P1 Q13

2025 HCI P1 Q13

Solution:

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