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The following is the prompt I am working on:

Suppose you have an investment account which grows by 10% every year, and you want to, in perpetuity, withdraw an increasing amount of money from this account, starting with 1 dollar at the end of the first year, 2 dollars at the end of the second, 3 dollars at the end of the third, and so forth, withdrawing $n after n years.

a.) Taking a0 to be the initial balance in this account, produce a linear nonhomogeneous recurrence to describe the balance after n years (note that the nth year involves both increasing by the interest earned, and decreasing by the size of the withdrawal).

For the problem I got the following answer

$$B_n=(1.1*B_{n-1}) -n$$

b)Solve the recurrence above, putting the answer in terms of the initial balance a0.

This is where I am struggling. I believe this is a nonhomogeneous recurrence relation that I need to find. But in almost all the examples I have worked so far, the non homogeneous part has a exponent. The issue could also be that the answer to a is incorrect, which would make any attempt at b incorrect.

2 Answers2

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Your recurrence looks correct. Let's solve it.

Suppose $a_n = b_n + c_n$, where $b_n$ is the solution to the homogeneous recurrence $-1.1 b_{n - 1} + b_n = 0$ and $c_n$ is a "particular" solution to $-1.1 c_{n - 1} + c_n = -n$.

The solution to the homogeneous problem is simple: $b_n = 1.1^n b_0$.

Hypothesize that $c_n = k_0 + k_1 n$; substituting, we obtain

$$-1.1 (k_0 + k_1 (n - 1)) + (k_0 + k_1 n) = -0.1 k_0 + 1.1 k_1 - 0.1 k_1 n = -n$$

which yields the linear system

$$\begin{cases} -0.1 k_0 + 1.1 k_1 = 0 \\ -0.1 k_1 = -1 \end{cases}$$

solved by $k_1 = 10$ and $k_0 = 110$.

So the general solution is

$$a_n = 110 + 10 n + 1.1^n b_0$$

with initial condition $a_0$ given. This means $b_0 = a_0 - 110$. So

$$\boxed{a_n = 110 + 10 n + 1.1^n (a_0 - 110)}$$

is the desired general solution.

K. Jiang
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1

You can also solve for $B_n$ by substitution.

$$\begin{align*} B_n &= \frac{11}{10} B_{n-1} - n \\ &= \frac{11}{10} \left(\frac{11}{10} B_{n-2} - (n-1)\right) - n \tag1 \\ &= \left(\frac{11}{10}\right)^2 B_{n-2} - (n + (n-1)) \\ &= \left(\frac{11}{10}\right)^2 \left(\frac{11}{10} B_{n-3} - (n-2)\right) - (n + (n-1)) \tag2\\ &= \left(\frac{11}{10}\right)^3 B_{n-3} - (n + (n-1) + (n-2)) \\ &~~ \vdots \end{align*}$$

After $(k)$ such rounds of substitution, you end up with

$$B_n = \left(\frac{11}{10}\right)^{k+1} B_{n-(k+1)} - \sum_{i=0}^k (n-i)$$

When $k=n-1$, you have

$$B_n = \left(\frac{11}{10}\right)^n B_0 - \sum_{i=0}^{n-1}(n-i) = \left(\frac{11}{10}\right)^n B_0 - \sum_{i=1}^n i$$

user170231
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