Mr. Chan will deposit $25000 in a bank account at the beginning of each year from 2007 to 2012. If the interest is compounded yearly and the interest rate is 4% p.a., find the total amount that he can get at the beginning of 2013. (corr. to 3 sig. fig.)

- anonymous

- schrodinger

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- anonymous

- ganeshie8

looks like annuity setup

- ganeshie8

P = 25,000
r = .04
n = 6

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## More answers

- ganeshie8

\(A = P[ (1+r)^n + (1+r)^{n-1} + .... (1+r)]\)
= \(P[\frac{1-(1+r)^n }{ 1-(1+r)}]\)

- ganeshie8

plug and chug..

- anonymous

ahhh....wait

- ganeshie8

ha take ur time not an easy problem :)

- primeralph

|dw:1372927372461:dw|

- anonymous

errr...my calculator shows that it's 165824.3866

- primeralph

@ganeshie8 Why the long stuff?

- ganeshie8

every year he is adding 25 grand

- anonymous

@primeralph this's a question about sequence

- primeralph

Well, both ways work.
Mine is just shorter.

- anonymous

@primeralph but your equation does match the answer, neither @ganeshie8 ..

- primeralph

Uh hold on.

- ganeshie8

prime i used ur formula and took a sum to arrive at the final formula

- primeralph

@kryton1212 What's the answer?

- anonymous

172000 after corr. to 3 sig fig

- anonymous

uhh?

- anonymous

i mean the correct answer..

- primeralph

It is not possible to get that much money after just 6 years.

- ganeshie8

http://www.wolframalpha.com/input/?i=25000%281-%281%2B.04%29%5E6%29%2F%281-%281%2B.04%29%29

- ganeshie8

im getting 165,824 @kryton1212
prime read the q again

- anonymous

i don't know how to do...
@hartnn show yourself-_-

- ganeshie8

which formula wont work ?

- primeralph

ohhh..........beginning of each year.
Hold on then.

- anonymous

i also think that @ganeshie8 's formula is correct...maybe i should ask my teacher about this question.
Let's move on to the next question that you both will be tagged by me a moment later

- anonymous

...

- primeralph

Hold on.

- ganeshie8

okie

- anonymous

maybe @primeralph goes on

- primeralph

Uh well, tag away if you're satisfied.

- anonymous

uh..

- ganeshie8

move on to new problem, im pretty sure what we worked is correct. if you do manually 6 times also, you would get the same answer.

- anonymous

at the beginning of 2013..................

- anonymous

@primeralph do you have any other ideas?

- primeralph

To check, you can do each step. If you get the same thing, let it go.

- anonymous

i am lazy XD...i will ask my teacher about this question tonight.
thank you both

- ganeshie8

np :)

- hartnn

i am getting
172457.362022

- anonymous

ohh, how?

- ganeshie8

3 sig figs, thats the right answer also... how ?

- hartnn

i calculated for each year...6 calculations :\

- hartnn

2008---25k(1+0.04)=26k
2009---(26k+25k)(1+0.04)=53040
2010---(53040+25k)(1+0.04)=81161.1
2011---(81161.1+25k)(1+0.04)=110408.064
2012---(110408.064+25k)(1+0.04)=140824.38
2013---(140824.38+25k)(1+0.04)=172457....

- anonymous

wow

- ganeshie8

oh every year rounding off to 3 sigfigs is it

- anonymous

that means we use the wrong method @ganeshie8 .....

- ganeshie8

may be

- anonymous

@hartnn million thanks. but how to present it in a shorter form...?

- hartnn

thats what i am thinking...
if we have it for just 2 years,
\(P(1+r)+ [P(1+r)+P](1+r) = P [(1+r)+(1+r)(2+r)]\)
so, maybe for 6 years, the formula will become
\(P (1+r)[1+(2+r)[1+(3+r)[1+(4+r)[1+(5+r)[1+(6+r)]]]]]\)
maybe there's a simplification involved, maybe...

- anonymous

wow, i think this's more complicated..

- hartnn

hmm.....maybe they wanted us to calculate each year's separately...because i don't see any explicit formula for this...

- anonymous

uhh, anyway, thank you @hartnn

- hartnn

hmm....welcome ^_^

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