Chapter 13.FOM, Problem 1P

Retirement Accounts Many college professors keep retirement savings with TIAA, the largest annuity program in the world. Interest on these accounts is compounded and credited daily. Professor Brown has $\$275,000$ on deposit with TIAA at the start of 2015 and receives $3.65\%$ interest per year on his account.

(a) Find a recursive sequence that models the amount $A_n$ in his account at the end of the $n\text{th}$ day of $2015$.

(b) Find the first eight terms of the sequence $A_n$, rounded to the nearest cent.

(c) Find a formula for $A_n$.

To determine

(a)

To find:

A recursive sequence that models the amount $A_n$ in account at the end of the $n\text{th}$ day of $2015$.

Answer to Problem 1P

Solution:

The recursive sequence that models the amount $A_n$ in account at the end of the $n\text{th}$ day of $2015$ is $A_n=1.0001\times A_{n-1}$

Explanation of Solution

Given:

The original deposit amount is $\$275000$ and interest rate is $3.65\%$ per year.

Interest is compounded and credited daily.

Approach:

Since the interest is compounded and credited daily, the interest rate for each day is:

$\left(\frac{3.65}{365}\right)\%=0.01\%$

The formula hence used will be:

$\left(\text{Amount at the end of}\text{day}\right)=\left(\text{Amount at the end of}theprevious\text{day}\right)+0.0001\times \left(\text{Amount at the end of the previous day}\right)$

Generalize the above formula for $n$ days to obtain the following formula:

$\left(\text{Amount at the end of }nth\text{day}\right)=1.0001\times \left(\text{Amount at the end of the previous day}\right)$

Calculation:

As explained above,

$\left(\text{Amount at the end of }nth\text{day}\right)=1.0001\times \left(\text{Amount at the end of the previous day}\right)$

If $A_n$ is the amount at the end of the $n\text{th}$ day, then the recursive formula becomes:

$A_n=\left(1.0001\right)\times A_{n-1},A_0=\$275000$

Therefore, a recursive sequence that models the amount $A_n$ in account at the end of the $n\text{th}$ day of $2015$ is $A_n=\left(1.0001\right)\times A_{n-1},A_0=\$275000$.

Conclusion:

Hence, a recursive sequence that models the amount $A_n$ in account at the end of the $n\text{th}$ day of $2015$.is $A_n=1.0001\times A_{n-1},A_0=\$275000$.

To determine

(b)

To find:

The first eight terms of the sequence $A_n$.

Answer to Problem 1P

Solution:

The first eight terms of the sequence $A_n$ are:

$\begin{array}{rll}{A}_0& =& 275000.00\\ A_1& =& 275027.50\\ A_2& =& 275055.00\\ A_3& =& 275082.51\end{array}$

$\begin{array}{rll}{A}_4& =& 275110.02\\ A_5& =& 275137.53\\ A_6& =& 275165.04\\ A_7& =& 275192.56\end{array}$

Explanation of Solution

Given:

The initial amount is $275000$.

Approach:

Use recursive relation calculated in part $\left(a\right)$.

Calculation:

From part $\left(a\right)$, the recursive formula is:

$A_n=\left(1.0001\right)\times A_{n-1},A_0=\$275000$ ……$\left(1\right)$

Substitute $n=1$, $2$, $3$, $4$, $5$, $6$, $7$ and $8$ successively in formula $\left(1\right)$,

Therefore, the first eight terms of the sequence $A_n$ are:

$\begin{array}{rll}{A}_0& =& 275000.00\\ A_1& =& 1.0001\times 275000\\ & =& 275027.50\end{array}$

$\begin{array}{rll}{A}_2& =& 1.0001\times 275027.50\\ & =& 275055.00\\ A_3& =& 1.0001\times 275055.00\\ & =& 275082.51\end{array}$

$\begin{array}{rll}{A}_4& =& 1.0001\times 275082.51\\ & =& 275110.02\\ A_5& =& 1.0001\times 275110.02\\ & =& 275137.53\end{array}$

$\begin{array}{rll}{A}_6& =& 1.0001\times 275137.53\\ & =& 275165.04\\ A_7& =& 1.0001\times 275165.04\\ & =& 275192.56\end{array}$

Conclusion:

Hence, the first eight terms of the sequence $A_n$ are:

$\begin{array}{rll}{A}_0& =& 275000.00\\ A_1& =& 275027.50\\ A_2& =& 275055.00\\ A_3& =& 275082.51\end{array}$

$\begin{array}{c}{A}_4=275110.02\\ A_5=275137.53\\ A_6=275165.04\\ A_7=275192.56\end{array}$

To determine

c)

To find:

A formula for $A_n$.

Answer to Problem 1P

Solution:

A formula for $A_n$ is $A_n=275000{\left(1.0001\right)}^n$.

Explanation of Solution

Given:

The initial amount is $275000$.

Approach:

Use part (b)

Calculation:

From the pattern,

$\begin{array}{rll}{A}_0& =& 275000.00\\ & =& {\left(1.0001\right)}^0\times 275000\\ A_1& =& 275027.50\\ & =& {\left(1.0001\right)}^1\times 275000\end{array}$

$\begin{array}{rll}{A}_2& =& {\left(1.0001\right)}^2\times 275000\\ A_3& =& {\left(1.0001\right)}^3\times 275000\\ A_4& =& {\left(1.0001\right)}^4\times 275000\\ A_5& =& {\left(1.0001\right)}^5\times 275000\end{array}$

$\begin{array}{rll}{A}_6& =& {\left(1.0001\right)}^6\times 275000\\ A_7& =& {\left(1.0001\right)}^7\times 275000\end{array}$

Therefore, a formula for $A_n$ is $A_n=275000{\left(1.0001\right)}^n$.

Conclusion:

Hence, a formula for $A_n$ is $A_n={\left(1.0001\right)}^n\times A_{n-1}$.