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It is a well-known fact the sum of counting numbers from 1 to 100 adds up to=5050. Now, suppose that these numbers are dollars deposited into an investment account at a rate of 5% compounded annually, in this manner:

Year 1=$1, year 2=$2, year 3=$3.......and so on till 100th year=$100. What would the total sum be in 100 years. P.S. My teacher says there is a short cut to this calculation!. Any help would be great. Thanks.

 Jan 21, 2016
 #1
avatar+129847 
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Assuming that the deposits are made at the beginning of each year ....we have....

 

1(1.05)^100  + 2(1.05)^99  + 3(1.05)^98  +   ...... +  99(1.05)^2  + 100(1.05)

 

n = 99

∑         (n + 1)(1.05)^(100-n)       =  about $55,451.05

n = 0

 

Maybe heureka knows a shortcut ???

 

 

 

 

cool cool cool

 Jan 21, 2016
 #2
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P.S. The deposits are made at the end of each year.

 Jan 21, 2016
 #3
avatar+129847 
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OK....assuming end of year deposits, we have

 

1(1.05)^99 + 2(1.05)^98  + 3(1.05)^97 +  ....+  99(1.05)^1 + 100(1.05)^0

 

n = 100

∑      (n)(1.05)^(100-n)  ≈  $52,810.53

n = 1

 

 

 

cool cool cool

 Jan 21, 2016
 #4
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Very good effort CPhill, but the answer in my book is different by a few thousands dollars. This might be up in "heureka's alley!". I shall get the "shortcut" from our teacher tomorrow.

 Jan 21, 2016
 #5
avatar+129847 
0

OK...sorry....I did my best......if heureka doesn't provide us with an answer, why don't you post the shortcut on here....????......

 

Thanks, CPhill

 

 

cool cool cool

 Jan 21, 2016
 #6
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Sorry, CPhill: After my 4th. comment above, I went to bed! It turns out that the answer is surprisingly simple. It is a relatively common and well-known TVM formula widely used in advanced accounting. And it is this:

FV= G {[1 + R]^N - RN - 1} / R^2=

FV= G {[ 1 + .05]^100 -(.05*100) - 1} / .05^2. Where:

G is a steadily increasing payment amount, that starts at G and increases by G for each subsequent period.

FV= 1{ 131.50125... - 5 - 1} / 0.0025

FV= {125.50125} / .0025

FV=$50,200.50.

And that is the answer in the book. Great effort on your part though, and I thank you.

 Jan 21, 2016

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