## kryton1212 2 years ago 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.)

1. kryton1212

@hartnn

2. ganeshie8

looks like annuity setup

3. ganeshie8

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

4. ganeshie8

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

5. ganeshie8

plug and chug..

6. kryton1212

ahhh....wait

7. ganeshie8

ha take ur time not an easy problem :)

8. primeralph

|dw:1372927372461:dw|

9. kryton1212

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

10. primeralph

@ganeshie8 Why the long stuff?

11. ganeshie8

every year he is adding 25 grand

12. kryton1212

@primeralph this's a question about sequence

13. primeralph

Well, both ways work. Mine is just shorter.

14. kryton1212

15. primeralph

Uh hold on.

16. ganeshie8

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

17. primeralph

18. kryton1212

172000 after corr. to 3 sig fig

19. kryton1212

uhh?

20. kryton1212

21. primeralph

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

22. ganeshie8
23. ganeshie8

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

24. kryton1212

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

25. ganeshie8

which formula wont work ?

26. primeralph

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

27. kryton1212

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

28. kryton1212

...

29. primeralph

Hold on.

30. ganeshie8

okie

31. kryton1212

maybe @primeralph goes on

32. primeralph

Uh well, tag away if you're satisfied.

33. kryton1212

uh..

34. 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.

35. kryton1212

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

36. kryton1212

@primeralph do you have any other ideas?

37. primeralph

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

38. kryton1212

39. ganeshie8

np :)

40. hartnn

i am getting 172457.362022

41. kryton1212

ohh, how?

42. ganeshie8

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

43. hartnn

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

44. 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....

45. kryton1212

wow

46. ganeshie8

oh every year rounding off to 3 sigfigs is it

47. kryton1212

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

48. ganeshie8

may be

49. kryton1212

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

50. 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...

51. kryton1212

wow, i think this's more complicated..

52. hartnn

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

53. kryton1212

uhh, anyway, thank you @hartnn

54. hartnn

hmm....welcome ^_^