T 4–4 Deposits needed to accumulate a future sum

  • Context:
  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Future Sum
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
T 4–4 Deposits needed to accumulate a future sum Judi wishes to accumulate \$8,000 by the end of 5 years by making equal annual end-of-year deposits over the next 5 years. If Judi can earn 7% on her investments, how much must she deposit at the end of each year to meet this goal?

$$\displaystyle A=P\left(1+\frac{r}{n}\right)^{nt}$$

ok not sure how plug this in

this complicated by the deposit made at the end of each year
 
Last edited:
Physics news on Phys.org
Let's let $D$ be the amount of the end of year deposits, and $i$ be the annual interest rate. So, the amount of the account at the end of year $n$ can be given by the difference equation:

$$A_n-(1+i)A_{n-1}=D$$ where $n\in\mathbb{N}$

The homogeneous solution is given by:

$$h_n=k_1(1+i)^n$$

And the particular solution is:

$$p_n=k_2$$

Plugging this into our difference equation, we find:

$$k_2-(1+i)k_2=D\implies k_2=-\frac{D}{i}$$

And so the closed form for $A_n$ is given by:

$$A_n=k_1(1+i)^n-\frac{D}{i}$$

Since:

$$A_1=D$$

We find:

$$k_1=\frac{D}{i}$$

And so the closed-form for $A_n$ is

$$A_n=\frac{D}{i}\left((1+i)^n-1\right)$$

Solve for $D$:

$$D=\frac{A_ni}{(1+i)^n-1}$$
 
good grief,,😎