Application of series and sequences question (1 Viewer)

[jimmysmith560]

Hello,

I have been facing an issue with this question and it would be great if someone helps:

Kevin pays $50 into a superannuation fund at the beginning of every month. The compound interest paid on the fund was originally 0.7% per month, but at the end of five years the rate increased to 0.8% per month. What is the total investment worth at the end of twelve years?

The answer is $13304 (to the nearest dollar), but I have no clue how to get it.

Thank you

 

[jskeza]

Ok let's take note of this firstly:
0.7 = (1+(0.7/100)) = 1.007
0.8 = 1.008 (similarly)
5 years = 60 months
12-5 = 7 years = 84 months

Now we will have two series

Our First Series:
A1 = 50(1.007)
A2 = 50(1.007)^2 + 50(1.007)
A3 = 50(1.007)^3 + 50(1.007)^2 + 50(1.007)
.
.
.
A60 = 50(1.007)^60 + 50(1.007)^59 + ... + 50(1.007)
= 50[ (1.007)^60 + (1.007)^59 + (1.007)^58 + ... + (1.007)]
Then using summation formula of an arithmetic series
= 50 [((1.007)(((1.007)^60)-1))/0.007]
= $3738.39

Our Second Series:
A1 = $3738.39(1.008) + 50(1.008)
A2 = $3738.39(1.008)^2 + 50(1.007)^2 + 50(1.008)
A3 = $3738.39(1.008)^3 + 50(1.007)^3 + 50(1.008)^2 + 50(1.008)
.
.
.
A84 = $3738.39(1.008)^84 + 50(1.008)^84 + 50(1.008)^83 + ... + 50(1.008)
= $3738.39(1.008)^84 + 50[ (1.008)^84 + (1.008)^83 + (1.008)^82 + ... + (1.008)]
Then using summation formula of an arithmetic series
= $3738.39(1.008)^84 + 50 [((1.008)(((1.008)^84)-1))/0.008]
= $13304.18362
= $13304 (nearest dollar)

Try leaving the rounding till the end (in first series)
Sorry for the crap formatting but I hope it helps.

 

[jimmysmith560]

Thanks a lot!