Jake invests $500 into a superannuation fund at the start of every year where it earns 6% per annum compounded monthly. Assume that there is an amount Bn in the fund after n years.
a) show that:
1) B1=500(1.005)^12
2)B2=500(1.005)^24 + 500(1.005)^12
3)B3=500(1.005)^36 + 500(1.005)^24 +500(1.005)^12
b) hence determine an expression for B20
c)jake plans to retire in 20 years. How much money will he have at this point in time
Please help with steps. I have been stuck in the algebra conversion from year to month in 1st question
?2014-08-02T02:54:52Z
So if it is 6% annual, then the interest is 0.06 times the principal. But it compounds monthly. There are 12 months in a year, so spread the rate evenly over 12 months (divide by 12).
0.06 / 12 = 0.005
So after 1 month, you have B[1 month] = 500 + 0.005*500, and factor out the 500, you have 500(1 + 0.005)
After 2 months you have B[2 month] = B[1month] + 0.005*B[1month], factor out B[1month], you have: B[1month](1 + 0.005), then substitute the value for B[1month]:
Do you see a pattern? B[m months] = 500(1 + 0.005)^m, where m is the number of months.
So if you have at the end of 1 year {12 months}, then B[1 year] = 500(1 + 0.005)^12.
The next day {beginning of second year} he deposits another 500.
So after 12 months, this money will be 500(1 + 0.005)^12, and the original money has been earning for 24 months, now, so it is 500(1 + 0.005)^24.
Add them together to get 500(1 + 0.005)^24 + 500(1 + 0.005)^12
Continue this thought process, the day after end of year 2, deposit another 500.
So at the end of year 3, the original 500 has earned for 36 months, the 500 deposited at the beginning of year 2 has earned for 24 months, and the 500 at start of year 3 has earned for 12 months, then you get the expression #3.
Lets look at these exponents. First note that we can factor out a 500, so we have: 500( (1.005)^36 + (1.005)^24 + (1.005)^12 )
Look at the first term and rewrite it as ((1.005)^12 ) * ((1.005)^12 ) * ((1.005)^12 ) = ((1.005)^12 )^3
Similarly the next term is ((1.005)^12 )^2. Let r = ((1.005)^12 ), so we have
500( r^3 + r^2 + r^1), do you see a pattern? For n years, we get 500(r^1 + r^2 + r^3 + r^4 + .... + r^n).
This is a Geometric Series. There is a formula for the sum.
Sum for this series is a( (1 - r^n) / (1 - r) ) {see the link}. The ratio between the terms is r, and also the first term is r, and the whole thing multiplied by 500, so we have:
500*r*( (1 - r^n) / (1 - r) ) ; remember that r = ((1.005)^12 ), and then they want 20 years so n = 20.