Finding Recurrence relations for combinatorics problems

Question

After you graduate you accept a job that promises a starting salary of $40,000$ and a raise at the end of each year equal to $5\%$ of your current salary plus $1000$. For example, your raise at the end of the first year is $3000$. Let $S_n$ be your salary after $n$ years, so that $S_0 = 40,000$.

A- Find a recurrence relation.

B- Determine how much you will be making after $2$ years, after $5$ years, after $10$ years.

I did part A. I don't know if I did it correctly, so it's $S_{n+1} = S_n \cdot 1.05+1000$ with $S_0=40,000$.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &S_{n} = 1.05\,S_{n - 1} + 1000 \implies S_{n} + 20000 = 1.05\pars{S_{n - 1} + 20000} \\[5mm] & \implies \bbx{\ds{S_{n} = 1.05^{\,n}\pars{S_{0} + 20000} - 20000} = 20000\pars{3 \times 1.05^{\,n} - 1}} \end{align}

$\ds{S_{2} = 46,150\,;\quad S_{5} = 56,576.9\,;\quad S_{10} = 77,733.7}$.

Answer 2

The recurrence relation \begin{align*} S_0 & = 40,000\\ S_{n + 1} & = S_n \cdot 1.05 + 1000 \end{align*} that you stated is correct.

Let's look at the first few terms of the sequence. \begin{align*} S_1 & = S_0 \cdot 1.05 + 1000\\ & = 40,000 \cdot 1.05 + 1000\\ S_2 & = S_1 \cdot 1.05 + 1000\\ & = (40,000 \cdot 1.05 + 1000) \cdot 1.05 + 1000\\ & = 40,000 \cdot 1.05^2 + 1000 \cdot 1.05 + 1000\\ S_3 & = S_2 \cdot 1.05 + 1000\\ & = (40,000 \cdot 1.05^2 + 1000 \cdot 1.05 + 1000) \cdot 1.05 + 1000\\ & = 40,000 \cdot 1.05^3 + 1000 \cdot 1.05^2 + 1000 \cdot 1.05 + 1000\\ & = 40,000 \cdot 1.05^3 + 1000(1.05^2 + 1.05 + 1) \end{align*} If we use the geometric series formula $$\sum_{k = 0}^{n} r^{k - 1} = 1 + r + r^2 + \cdots r^{n - 1} = \frac{1 - r^n}{1 - r}$$ we can express $S_3$ in the form $$S_3 = 40,000 \cdot 1.05^3 + 1000 \cdot \frac{1 - 1.05^3}{1 - 1.05}$$ Can you find a formula for $S_n$?