   Chapter 13.FOM, Problem 1P ### Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742

#### Solutions

Chapter
Section ### Algebra and Trigonometry (MindTap ...

4th Edition
James Stewart + 2 others
ISBN: 9781305071742
Textbook Problem

# 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 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 An in account at the end of the nth day of 2015. Answer Solution: The recursive sequence that models the amount An in account at the end of the nth day of 2015 is An=1.0001×An1 Explanation 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:

(3.65365)%=0.01%

The formula hence used will be:

(Amount at the end of day)=(Amount at the end of the previous day)+0.0001×(Amount at the end of the previous day)

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

(Amount at the end of nth day)=1.0001×(Amount at the end of the previous day)

Calculation:

As explained above,

(Amount at the end of nth day)=1.0001×(Amount at the end of the previous day)

If An is the amount at the end of the nth day, then the recursive formula becomes:

An=(1.0001)×An1, A0=$275000 Therefore, a recursive sequence that models the amount An in account at the end of the nth day of 2015 is An=(1.0001)×An1, A0=$275000.

Conclusion:

Hence, a recursive sequence that models the amount An in account at the end of the nth day of 2015.is An=1.0001×An1, A0=$275000. To determine (b) To find: The first eight terms of the sequence An. Answer Solution: The first eight terms of the sequence An are: A0=275000.00A1=275027.50A2=275055.00A3=275082.51 A4=275110.02A5=275137.53A6=275165.04A7=275192.56 Explanation Given: The initial amount is 275000. Approach: Use recursive relation calculated in part (a). Calculation: From part (a), the recursive formula is: An=(1.0001)×An1, A0=$275000 ……(1)

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

Therefore, the first eight terms of the sequence An are:

A0=275000.00A1=1.0001×275000=275027.50

A2=1.0001×275027.50=275055.00A3=1.0001×275055.00=275082.51

A4=1.0001×275082.51=275110.02A5=1.0001×275110.02=275137.53

A6=1.0001×275137.53=275165.04A7=1.0001×275165.04=275192.56

Conclusion:

Hence, the first eight terms of the sequence An are:

A0=275000.00A1=275027.50A2=275055.00A3=275082.51

A4=275110.02A5=275137.53A6=275165.04A7=275192.56

To determine

c)

To find:

A formula for An.

Solution:

A formula for An is An=275000(1.0001)n.

Explanation

Given:

The initial amount is 275000.

Approach:

Use part (b)

Calculation:

From the pattern,

A0=275000.00=(1.0001)0×275000A1=275027.50=(1.0001)1×275000

A2=(1.0001)2×275000A3=(1.0001)3×275000A4=(1.0001)4×275000A5=(1.0001)5×275000

A6=(1.0001)6×275000A7=(1.0001)7×275000

Therefore, a formula for An is An=275000(1.0001)n.

Conclusion:

Hence, a formula for An is An=(1.0001)n×An1.

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