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"This is the "teacher's guide" book in Kenny Felder's "Advanced Algebra II" series. This text was created with a focus on 'doing' and 'understanding' algebra concepts rather than simply hearing […]"

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# Extra Credit

Module by: Kenny M. Felder. E-mail the author

Summary: An extra problem on compound interest to be used for extra credit. This module is part of the Teacher's Guide.

## An extra cool problem you may want to use as an extra credit or something

### Exercise 1

A bank gives i % i% interest, compounded annually. (For instance, if i = 6 i=6, that means 6% interest.) You put A A dollars in the bank every year for n n years. At the end of that time, how much money do you have?

#### Note:

(The fine print: Let’s say you make your deposit on January 1 every year, and then you check your account on December 31 of the last year. So if n = 1 n=1, you put money in exactly once, and it grows for exactly one year.)

#### Solution

The money you put in the very last year receives interest exactly once. “Receiving interest” in a year always means being multiplied by 1+i1001+i100 size 12{ left (1+ { {i} over {"100"} } right )} {}. (For instance, if you make 6% interest, your money multiplies by 1.06.) So the A A dollars that you put in the last year is worth, in the end, A 1+i100A1+i100 size 12{ left (1+ { {i} over {"100"} } right )} {}.

The previous year’s money receives interest twice, so it is worth A 1+i1002A1+i1002 size 12{ left (1+ { {i} over {"100"} } right ) rSup { size 8{2} } } {} at the end. And so on, back to the first year, which is worth A 1+i100nA1+i100n size 12{ left (1+ { {i} over {"100"} } right ) rSup { size 8{n} } } {} (since that initial contribution has received interest n n times).

So we have a Geometric series:

S = A 1+i100+ A 1+i1002+ ... + A 1+i100nS=A1+i100 size 12{ left (1+ { {i} over {"100"} } right )} {}+A1+i1002 size 12{ left (1+ { {i} over {"100"} } right ) rSup { size 8{2} } } {}+...+A1+i100n size 12{ left (1+ { {i} over {"100"} } right ) rSup { size 8{n} } } {}

We resolve it using the standard trick for such series: multiply the equation by the common ratio, and then subtract the two equations.

1+i100S = A 1+i1002+ ... + A 1+i100n+ A 1+i100n+11+i100 size 12{ left (1+ { {i} over {"100"} } right )} {}S=A1+i1002 size 12{ left (1+ { {i} over {"100"} } right ) rSup { size 8{2} } } {}+...+A1+i100n size 12{ left (1+ { {i} over {"100"} } right ) rSup { size 8{n} } } {}+A1+i100n+1 size 12{ left (1+ { {i} over {"100"} } right ) rSup { size 8{n+1} } } {}

S = A 1+i100+ A 1+i1002 + ... + A 1+i100nS=A1+i100 size 12{ left (1+ { {i} over {"100"} } right )} {}+A1+i1002 size 12{ left (1+ { {i} over {"100"} } right ) rSup { size 8{2} } } {}+...+A1+i100n size 12{ left (1+ { {i} over {"100"} } right ) rSup { size 8{n} } } {}

i100S = A 1+i100n+1 A 1+i100i100 size 12{ left ( { {i} over {"100"} } right )} {}S=A1+i100n+1 size 12{ left (1+ { {i} over {"100"} } right ) rSup { size 8{n+1} } } {}A1+i100 size 12{ left (1+ { {i} over {"100"} } right )} {}

S = 100AiS=100Ai size 12{ { {"100"A} over {i} } } {}1+i100n+11+i1001+i100n+11+i100 size 12{ left [ left (1+ { {i} over {"100"} } right ) rSup { size 8{n+1} } - left (1+ { {i} over {"100"} } right ) right ]} {}

### Example 1

Example: If you invest $5,000 per year at 6% interest for 30 years, you end up with: 100(5000)6[ 1.0631 1.06 ] =$ 419,008.39 100(5000)6 size 12{ { {"100" $$"5000"$$ } over {6} } } {}[1.06311.06]=$419,008.39 Not bad for a total investment of$150,000!

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