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Compound interest

daigo

Member
Jun 27, 2012
60
A deposit of $$25 is made at the beginning of the 1st month, and successive monthly deposits after that is $25 more than the previous month (2nd month is a $50 deposit, 3rd month is a $75, etc.). At the beginning of the next year (after 12 months), the deposit cycle is reset back to $25 the first month, etc. and this pattern continues for 5 years. The account pays 5% compounded interest monthly at the end of each month. What is the balance of the account after 5 years?
So I'm trying to find out the formula by writing out the expanded version first and simplifying it, but I'm not sure how to write this in terms of exponents.

Month 1: [tex]25 + 25(.05)[/tex]

Let x = [tex]25 + 25(.05)[/tex]

Month 2: [tex](x + 50)(.05) + (x + 50)[/tex]

Month 3: [tex]((x + 50)(.05) + (x + 50) + 75)(.05) + ((x + 50)(.05) + (x + 50) + 75)[/tex]

Month 4: (((x + 50)(.05) + (x + 50) + 100)(.05) + ((x + 50)(.05) + (x + 50) + 100))(.05) + ((x + 50)(.05) + (x + 50) + 100)(.05) + ((x + 50)(.05) + (x + 50) + 100)

But then I remembered the formula will probably change after 12 months since the deposits start over, but the balance is different...so I'm not sure how else to really approach this.
 
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daigo

Member
Jun 27, 2012
60
So far, this is what I got for the first year:

[tex](25m)(1 + \frac{.05}{12})(\frac{1 - (1 + \frac{.05}{12})^{12}}{{1 - (1 + \frac{.05}{12})}})[/tex]
where m = the month...

Not sure if this is correct, but do I need to make a new formula for each year to make up for the "new" principal in the account (the total from the preceding year)?
 
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mathpro1

New member
Aug 16, 2012
1
So I'm trying to find out the formula by writing out the expanded version first and simplifying it, but I'm not sure how to write this in terms of exponents.

Month 1: [tex]25 + 25(.05)[/tex]

Let x = [tex]25 + 25(.05)[/tex]

Month 2: [tex](x + 50)(.05) + (x + 50)[/tex]

Month 3: [tex]((x + 50)(.05) + (x + 50) + 75)(.05) + ((x + 50)(.05) + (x + 50) + 75)[/tex]

Month 4: (((x + 50)(.05) + (x + 50) + 100)(.05) + ((x + 50)(.05) + (x + 50) + 100))(.05) + ((x + 50)(.05) + (x + 50) + 100)(.05) + ((x + 50)(.05) + (x + 50) + 100)

But then I remembered the formula will probably change after 12 months since the deposits start over, but the balance is different...so I'm not sure how else to really approach this.

okay lets use the compound interest formula:

A=P(1+r/n)^nt

where A is the amount of the money in the account after the interest paid. P is the starting amount or the principle. R is the intrest rate, N is the number of times it will be compounded if monthly than it would be 12, And T is time its in the account.
For this problem since your adding money to the account you will need to do 12 calculations time the 5 years.
All that would change is the principle

Year 1:
Month 1: 25 (1+.05/12)^12*1= \$26.28
Month 2: (26.28+50) (1+.05/12)^12*1=\$80.18
Month 3: (80.18+75)*(1+.05/12)^12*1= \$163.12
Month 4: (163+100)*(1+.05/12)^12*1= \$276.58
Month 5: (276.58+125)*(1+.05/12)^12*1=\$422.13
Month 6: (422.13+150)*(1+.05/12)^12*1=\$601.40
Month 7: (601.40+175)*(1+.05/12)^12*1=\$816.12
Month 8: (816.12+200)*(1+.05/12)^12*1=\$1068.11
Month 9: (1068.11+225)*(1+.05/12)^12*1=\$1359.26
Month 10: (1359.26+250)*(1+.05/12)^12*1=\$1691.60
Month 11: (1691.60+275)*(1+.05/12)^12*1=\$2067.21
Month 12: (2067.21+300)*(1+.05/12)^12*1=\$2488.33

Total for year 1 is 2488.33

You would need to do this for all 5 years.

but dont forget that each new year the deposits go up 25 and start over each year.
 
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Wilmer

In Memoriam
Mar 19, 2012
376
okay lets use the compound interest formula:
A=P(1+r/n)^nt
where A is the amount of the money in the account after the interest paid. P is the starting amount or the principle. R is the intrest rate, N is the number of times it will be compounded if monthly than it would be 12, And T is time its in the account.
For this problem since your adding money to the account you will need to do 12 calculations time the 5 years.
All that would change is the principle
Year 1:
Month 1: 25 (1+.05/12)^12*1= \$26.28
Month 2: (26.28+50) (1+.05/12)^12*1=\$80.18
Month 3: (80.18+75)*(1+.05/12)^12*1= \$163.12
Month 4: (163+100)*(1+.05/12)^12*1= \$276.58
Month 5: (276.58+125)*(1+.05/12)^12*1=\$422.13
Month 6: (422.13+150)*(1+.05/12)^12*1=\$601.40
Month 7: (601.40+175)*(1+.05/12)^12*1=\$816.12
Month 8: (816.12+200)*(1+.05/12)^12*1=\$1068.11
Month 9: (1068.11+225)*(1+.05/12)^12*1=\$1359.26
Month 10: (1359.26+250)*(1+.05/12)^12*1=\$1691.60
Month 11: (1691.60+275)*(1+.05/12)^12*1=\$2067.21
Month 12: (2067.21+300)*(1+.05/12)^12*1=\$2488.33
Total for year 1 is 2488.33
You would need to do this for all 5 years.
That's too high; deposited is total of $1950; 2488.33 - 1950 = 534.33:
that's way too much interest! Your powers need to reduce by 1 each month.

The correct accumulation for 1 year is 1988.35; account "looks like":
Code:
    DEPOSIT    INTEREST  BALANCE
00   25.00       .00      25.00
01   50.00       .10      75.10
02   75.00       .32     150.42
....
10  275.00      5.80    1673.13
11  300.00      6.97    1980.10
12              8.25    1988.35
Next step is simply to use 1988.35 as 5 annual deposits
earning interest at 5% cpd monthly: ~5.116% annual.

Formula: F = D[(1 + i)^n - 1] / i
F = 1988.35(1.05116^5 - 1) / .05116 = 11012.38
 
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