McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
5. Recursion and Iteration
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Exercise 7 Page 695

Practice makes perfect
a Let a_n be the amount of money in a deposit at the start of the n^(th) month. Ben financed a $1500 rowing machine. His bill increased by 1 % at the end of each month. Furthermore, he makes a $100 payment each month. This tells us that a_1=1500.
a_n--- Money at the end of then^(th)month

Let's find a formula for a_n for n> 1. The balance at the start of the n^(th) month is a_(n-1). First, the interest is calculated. Since we add 1 %=0.01 of interest, we should multiply the actual amount by 1+0.01=1.01. Then we subtract 100 dollars and get the following equation. a_n=1.01 a_(n-1)- 100 Finally, we can write a recursive rule for a_n. Recursive Rule: a_1=1500, a_n=1.01 a_(n-1)- 100forn>1

b We are asked to find the balance owed at the end of each month. Therefore, we should find a_n for n=2, 3, 4, and 5. We will use the recursive formula from Part A.
n a_n=1.01a_(n-1)-100 1.01a_(n-1)-100 a_n
1 a_1=1500 - a_1= 1500
2 1.01a_1-100 1.01( 1500)-100 a_2= 1415
3 1.01a_2-100 1.01( 1415)-100 a_3= 1329.15
4 1.01a_3-100 1.01( 1329.15)-100 a_4=1242.44
5 1.01a_4-100 1.01(1242.44)-100 a_5=1154.86

This tells us that the balances are $ 1415, $ 1329.15, $1242.44, and $1154.86.

Extra

Calculations Using a Spreadsheet
Let's enter a_1=1500 in cell A1 and write =Round((1.01)*(A1)-100,2) in cell A2. When we hit enter, we will get the following.
Next, copy cell A2, highlight cells A3 through A5, and paste.
From the spreadsheet we can get that the balances are $1415, $1329.15, $1242.44, and $1154.86.
c We are to find how much interest has accumulated after the first six months. The interest after the n^(th) month is exactly 0.01a_n. Let's use the table from Part B, calculate a_6, and calculate the interest.
n a_n=1.01a_(n-1)-100 1.01a_(n-1)-100 a_n 0.01a_n
1 a_1=1500 - a_1= 1500 0.01( 1500)=15
2 1.01a_1-100 1.01( 1500)-100 a_2= 1415 0.01( 1415)=14.15
3 1.01a_2-100 1.01( 1415)-100 a_3= 1329.15 0.01( 1329.15)≈ 13.29
4 1.01a_3-100 1.01( 1329.15)-100 a_4=1242.44 0.01(1242.44)≈ 12.42
5 1.01a_4-100 1.01(1242.44)-100 a_5=1154.86 0.01(1154.86)≈ 11.55
6 1.01a_5-100 1.01(1154.86)-100 a_6= 1066.41 0.01( 1066.41)≈10.66

Now, to find the total interest during the first six months we will add the interest after each month. 15+14.15+13.29+12.42+11.55+10.66 = 77.07 Therefore, the total interest is about $77.07.

Extra

Calculations Using a Spreadsheet
Let's enter a_1=1500 in cell A1 and write =Round((1.01)*(A1)-100,2) in cell A2. When we hit enter we will get the following.
Next, copy cell A2, highlight cells A3 through A5, and paste.
Now, write =Round((0.01)*(A1),2) in cell B1. When we hit enter we will get the following.
Next, copy cell B1, highlight cells B2 through B6, and paste.
Now, enter Sum in cell A8 and write =Sum(B1:B6) in cell B8 to get the sum of all cells from cell B1 to B6 . When we hit enter, we will get the following.
Therefore, the total interest is about $77.07.