# Quick HELP GEOMETRIC SERIES SUM

1. Mar 11, 2005

### aisha

Help Geometric Sum Help!!! Plz

Hi here is the question It says a retired hockey star wants to set up a scholarship fund to assist an underpriveleged child who would like to go to a post secondary institution. He wants to ensure that the student will have $6000 per year for 5 years. HOw much should he give to the institution now? to ensure this can happen, if the institution is able to invest the money at 10%/a compounded annually? I was to set up a line diagram showing the present value of each of the$6000 payments which I did taking all of the present values and putting them in a geometric series I got

6000(1+0.10)^-1+6000(1+0.10)^-2+6000(1+0.10)^-3+6000(0.10)^-4+6000(0.10)^-5

Now I think the common ratio for this geometric series is r=1.1^-1 = 0.9091
and a=6000

using these two I pluged these values into the sn formula sn=[a(r^(n) -1)]/(r-1)

and I got my final sum of the geometric series to be $25,019.65 I dont know if this is correct can someone plz help me out plzzzzzzz I want to hand this in today if possible. Last edited: Mar 12, 2005 2. Mar 12, 2005 ### aisha Help Me Please Anyone Please!!! When adding the terms in the geometric series without a formula and using a graphing calculator I get the amount he should pay now to be$22,744.72, but how come if this value is correct I cant get it with the sn formula?

WHY WONT ANYONE HELP ME? 45 views yet no reply I've been waiting for sooo long PLZ HELP ME!!! Where are all the tutors HELP ME !

Last edited: Mar 12, 2005
3. Mar 12, 2005

### harsh

I am not exactly sure why you want to sum a geometric series. Summing a series is not usually a very easy thing to do. From your question, it seems that you can some sort of compound-interest formula to figure out the starting amount if you want to end up with 6,000 at the end of 5 years with 10% interest. It doesnt require a geometric series of any kind. I think you making the problem complicated than it should be.

Btw the way, your value of 22,000 cant be correct. It has to be much less than 6,000

- harsh

4. Mar 12, 2005

### aisha

I Need More Help Plz

5. Mar 12, 2005

At the end of 5 yrs,it should have the necesary 30,000$...Since the interest is compounded annually,the equation is $$x(1+0.1)^{5}=30,000$$ ,with the solution $$x\approx 18,628$$...(dollars) Daniel. P.S.That's what i made of your problem. 6. Mar 12, 2005 ### xanthym According to the problem statement, the fund is established at beginning of year #1 (at which time$6000 is immediately withdrawn for tuition) and remains in effect for 4 subsequent years (and $6000 is withdrawn each year at year's start). Each year between withdrawals, the fund accrues 10% interest on the remaining monetary value. Problem requires$0 to remain just after the year #5 withdrawal of $6000. Let D be the scholarship dollar value at beginning of year #1 when fund is established. Then the problem requires: ((((D - 6000)*(1.1) - 6000)*(1.1) - 6000)*(1.1) - 6000)*(1.1) - 6000 = 0 ::: ⇒ D*(1.1)4 - 6000*{(1.1)4 + (1.1)3 + (1.1)2 + (1.1) + 1} = 0 ::: ⇒ D*(1.1)4 - 6000*{(1.1)5 - 1}/{(1.1) - 1} = 0 ::: <--- Geometric Series Formula ::: ⇒ D*(1.4641) - 6000*{0.6105}/{0.1} = 0 ::: ⇒ D = 6000*{0.6105}/{(0.1)*(1.4641)} ::: ⇒ D = ($25,019)

~~

Last edited: Mar 13, 2005
7. Mar 13, 2005

### Data

A geometric series is an exception. Here we have

$$\sum_{i=1}^n r^i$$

Letting $$S_n = r + r^2 + r^3 + .\ .\ . + r^n$$ gives

$$\frac{S_n}{r} = 1 + r + r^2 + .\ .\ . + r^{n-1}$$

and thus

$$S_n\left(1-\frac{1}{r}\right) = r^n - 1 \Longrightarrow \sum_{i=1}^n r^i = S_n = \frac{r^n - 1}{1-\frac{1}{r}} = \frac{r^{n+1} - r}{r-1}$$

Last edited: Mar 13, 2005
8. Mar 13, 2005

### dextercioby

I had a hunch,i'd get this problem wrong...:tongue2:

Daniel.

9. Mar 13, 2005

### aisha

I got both answers 25,019.19 and 18,627.34 plus plenty more answers lol, but I have finally got the right answer from the online teacher tutors on ilc.org. The geometric series should look like this

6000(1+0.10)^-5+6000(1+0.10)^-4+6000(1+0.10)^-3+6000(0.10)^-2+6000(0.10)^-1

the common ratio r=1.1
n=5 from the question
and a=6000(1.1)^-5 this my teacher told me im still not sure y

anyways if u input these numbers into the formula sn=[a(r^(n)-1]/r-1

you will get {6000(1.1)^-5[(1.1)^(5)-1]}/all divided by {0.1}

im not sure how but the top is simplified to [6000-6000(1.1)^-5]/0.1 and when u plug this into the calculator the final value is \$22744.72 which is what u get if u simply add the terms given in the geometric series.

If anyone can explain how a=6000(1.1)^-5 and how to simplify the brackets in the sn formula plz do tell me thanks everyone for trying it was a toughy, but one of my shortest assignments ever :rofl:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook