MHB T 4–4 Deposits needed to accumulate a future sum

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To accumulate $8,000 in 5 years with annual deposits at a 7% interest rate, the formula for future value with regular deposits is applied. The difference equation for the account balance is established, leading to a closed-form solution for the amount after n years. The final formula for the annual deposit, D, is derived as D = (A_n * i) / ((1 + i)^n - 1). By substituting the values, one can calculate the required annual deposit. This approach effectively determines the necessary contributions to reach the savings goal.
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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
 
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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,,😎
 
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