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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.)

Mathematics
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looks like annuity setup
P = 25,000 r = .04 n = 6

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Other answers:

\(A = P[ (1+r)^n + (1+r)^{n-1} + .... (1+r)]\) = \(P[\frac{1-(1+r)^n }{ 1-(1+r)}]\)
plug and chug..
ahhh....wait
ha take ur time not an easy problem :)
|dw:1372927372461:dw|
errr...my calculator shows that it's 165824.3866
@ganeshie8 Why the long stuff?
every year he is adding 25 grand
@primeralph this's a question about sequence
Well, both ways work. Mine is just shorter.
@primeralph but your equation does match the answer, neither @ganeshie8 ..
Uh hold on.
prime i used ur formula and took a sum to arrive at the final formula
@kryton1212 What's the answer?
172000 after corr. to 3 sig fig
uhh?
i mean the correct answer..
It is not possible to get that much money after just 6 years.
http://www.wolframalpha.com/input/?i=25000%281-%281%2B.04%29%5E6%29%2F%281-%281%2B.04%29%29
im getting 165,824 @kryton1212 prime read the q again
i don't know how to do... @hartnn show yourself-_-
which formula wont work ?
ohhh..........beginning of each year. Hold on then.
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
...
Hold on.
okie
maybe @primeralph goes on
Uh well, tag away if you're satisfied.
uh..
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.
at the beginning of 2013..................
@primeralph do you have any other ideas?
To check, you can do each step. If you get the same thing, let it go.
i am lazy XD...i will ask my teacher about this question tonight. thank you both
np :)
i am getting 172457.362022
ohh, how?
3 sig figs, thats the right answer also... how ?
i calculated for each year...6 calculations :\
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....
wow
oh every year rounding off to 3 sigfigs is it
that means we use the wrong method @ganeshie8 .....
may be
@hartnn million thanks. but how to present it in a shorter form...?
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...
wow, i think this's more complicated..
hmm.....maybe they wanted us to calculate each year's separately...because i don't see any explicit formula for this...
uhh, anyway, thank you @hartnn
hmm....welcome ^_^

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