- Thread starter
- #1

A. P625

B. P1000

C. P1250

D. P1625

- Thread starter Joe_1234
- Start date

- Thread starter
- #1

A. P625

B. P1000

C. P1250

D. P1625

- Admin
- #2

I would let \(A_n\) be the amount in the account at the end of year \(n\), and \(W\) be the amount withdrawn at the end of year \(n\). Then we may model this problem with the following recursion:

\(\displaystyle A_{n+1}=1.05A_{n}-W\)

Can you give the homogeneous solution (based on the root of the characteristic equation)?

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- #3

\(\displaystyle A_{n+1}=1.025^2A_{n}-W=\left(\frac{41}{40}\right)^2A_{n}-W\)

And so our homogeneous solution will take the form:

\(\displaystyle h_n=c_1\left(\frac{41}{40}\right)^{2n}\)

Our particular solution will take the form:

\(\displaystyle p_n=B\)

Substitution into the recursion yields:

\(\displaystyle B-\left(\frac{41}{40}\right)^2B=-W\)

\(\displaystyle \left(\frac{9}{40}\right)^2B=W\implies B=\left(\frac{40}{9}\right)^2W\)

And so by the principle of superposition, we obtain:

\(\displaystyle A_{n}=h_n+p_n=c_1\left(\frac{41}{40}\right)^{2n}+\left(\frac{40}{9}\right)^2W\)

\(\displaystyle A_0=c_1+\left(\frac{40}{9}\right)^2W\implies c_1=A_0-\left(\frac{40}{9}\right)^2W\)

And so our closed form is:

\(\displaystyle A_{n}=\left(A_0-\left(\frac{40}{9}\right)^2W\right)\left(\frac{41}{40}\right)^{2n}+\left(\frac{40}{9}\right)^2W\)

To finish the problem, we want to set:

\(\displaystyle \lim_{n\to\infty}A_n=0\)

In order for that to happen (for the limit to be finite), we need:

\(\displaystyle W=\left(\frac{9}{40}\right)^2A_0\)

Using the value \(A_0=25000\), we then find:

\(\displaystyle W=\frac{10125}{8}=1265.625\)

- Thread starter
- #4

Tnx Sir

\(\displaystyle A_{n+1}=1.025^2A_{n}-W=\left(\frac{41}{40}\right)^2A_{n}-W\)

And so our homogeneous solution will take the form:

\(\displaystyle h_n=c_1\left(\frac{41}{40}\right)^{2n}\)

Our particular solution will take the form:

\(\displaystyle p_n=B\)

Substitution into the recursion yields:

\(\displaystyle B-\left(\frac{41}{40}\right)^2B=-W\)

\(\displaystyle \left(\frac{9}{40}\right)^2B=W\implies B=\left(\frac{40}{9}\right)^2W\)

And so by the principle of superposition, we obtain:

\(\displaystyle A_{n}=h_n+p_n=c_1\left(\frac{41}{40}\right)^{2n}+\left(\frac{40}{9}\right)^2W\)

\(\displaystyle A_0=c_1+\left(\frac{40}{9}\right)^2W\implies c_1=A_0-\left(\frac{40}{9}\right)^2W\)

And so our closed form is:

\(\displaystyle A_{n}=\left(A_0-\left(\frac{40}{9}\right)^2W\right)\left(\frac{41}{40}\right)^{2n}+\left(\frac{40}{9}\right)^2W\)

To finish the problem, we want to set:

\(\displaystyle \lim_{n\to\infty}A_n=0\)

In order for that to happen (for the limit to be finite), we need:

\(\displaystyle W=\left(\frac{9}{40}\right)^2A_0\)

Using the value \(A_0=25000\), we then find:

\(\displaystyle W=\frac{10125}{8}=1265.625\)