Formula for compounding interest with yearly increasing additional payments

Question

I would like to find a formula for the following:

Let's say Karen makes $75,000 annual salary and is currently investing 10% of her income every year. Each year, however, Karen receives a salary increase of 5%. She is investing for a total of 30 years (annual payments made at the end of each year). The return on investment is 12%.

I understand the compound formula with additional payments as:

$$FV = P(1+i)^n + A \frac{(1+i)^n - 1}{i}$$

FV = Future Value P = Principal i = Interest Rate n = Number of periods

"A" in the formula increases each year by 5% due to the 5% annual salary increase. I know that I can use this formula for each year and add the sum of all years for 30 years of investing to get my answer. However, I would like to know if there is a formula to get the end result without having to add up each year individually. Thanks!

Answer 1

Suppose we have a constant per period interest rate $i$ and have the following investment schedule:

$$ \text{period } 0: \text{ I invest } P,\\ \text{period } 1: \text{ I invest } A_1,\\ \vdots \\ \text{period } n: \text{ I invest } A_n.\\ $$

Then the future value of our investment as of period $n$ will be

$$P(1+i)^n+\sum_{k=1}^nA_k(1+i)^{n-k}.$$

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The above is general, but now let's look at the special case where the additional payments grow at a constant rate $g$, i.e. $$A_k=A_1(1+g)^{k-1},k\geq1 .$$ Then the above formula is a finite geometric series:

$$P(1+i)^n+\sum_{k=1}^nA_1(1+g)^{k-1}(1+i)^{n-k}\\ =P(1+i)^n+A_1(1+i)^{n-1}\sum_{k=0}^{n-1}\left(\frac{1+g}{1+i}\right)^{k}\\ =P(1+i)^n+A_1(1+i)^{n-1}\frac{\left(\frac{1+g}{1+i}\right)^{n}-1}{\frac{1+g}{1+i}-1}\\ =P(1+i)^n+A_1\frac{(1+i)^n-(1+g)^n}{i-g}, $$ which is valid if $i\neq g$. Note this boils down to the additional payments formula you mention when there is no growth in the additional payments: $g=0$.

Your problem would use this formula, with $P=75,000*10\%,A_1=75,000*10\%*(1.05),g=5\%,i=12\%,n=30.$

As an addendum, in the case $i=g$, you can work out the future value is $P(1+i)^n+nA_1(1+i)^{n-1}.$

Answer 2

In order to provide an overall formula for your (more complicated) problem, you have to go back to basics.

First, some simplifying assumptions, upon which my entire answer is based:

• Karen starts working on 2021-01-01, at $75,000$ a year.
• Karen makes her first deposit on 2021-12-31.
• Karen makes her last deposit on 2050-12-31.
• The question being asked is what the total value of the deposits will be on 2050-12-31.
• On 2021-12-31, Karen deposits $(75k) \times (0.1)$.
• On 2022-01-01, Karen's salary increases to $(75k) \times (1.05)$.
• On 2022-12-31, just before Karen makes her second deposit, her balance is $(75k) \times (0.1) \times (1.12).$

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Let $P = 75k \times (0.1)$.
Let $t = 1.12$.

Note that if a deposit $D$ is made at the end of year $1$, then at the end of year $30$ that deposit is worth $Dt^{(29)}.$

So, let $D_k$ denote the exact size of the deposit made at the end of year $k ~: ~k \in \{1,2,\cdots,30\}.$

Then, at the end of year 30, $D_k$, by itself, will have grown to $D_k \times t^{(30-k)}.$

Based on the assumptions in the first section of my answer:

$D_1 = P.$
$D_2 = P \times 1.05.$
$D_3 = P \times 1.05^2.$
$D_4 = P \times 1.05^3.$
$\cdots$
$D_{(29)} = P \times 1.05^{(28)}.$
$D_{(30)} = P \times 1.05^{(29)}.$

Therefore, on 2050-12-31, just after the $30$-th (i.e. the last) deposit, the balance will be

$$\sum_{k=1}^{30} \left[D_k \times t^{(30-k)}\right]$$

$$= \sum_{k=1}^{30} \left[P \times 1.05^{(k-1)} \times t^{(30-k)}\right].\tag1 $$

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In order to simplify the expression in (1) above, you (again) go back to basics.

$$1 + a + a^2 + \cdots + a^{(n-1)} = \frac{1 - a^n }{1 - a}.\tag2 $$

So, the challenge will be to re-write (1) above, so that it fits into the format of (2) above. So, $n$ will be set to $30$, the first term will be represented by $1$, the second term will be represented by $a^1$, and the $30$th term will be represented by $a^{(29)}$.

Also, this overall sum will be multiplied by a constant $C$.

In (1) above, the actual size of the first term is:

$\displaystyle P \times 1.05^{0} \times t^{(29)} = 75000 \times (0.1) \times (1) \times 1.12^{(29)}.$

Therefore,
let $\displaystyle C = 75000 \times (0.1) \times (1) \times 1.12^{(29)}.$

Now the question becomes what is the common ratio between terms in the expression in (1) above.

The common ratio is $\displaystyle \frac{1.05}{1.12}$.

Therefore, let $\displaystyle a = \frac{1.05}{1.12}.$

Then, with $C$ and $a$ as described, the formula in (2) above, may be applied. Therefore, the final value is

$$C \times \frac{1 - a^{(29)}}{1 - a}.$$