Getteral09 said:
i try doing it on 9 yrs but i can't seem to get the answer as Watatank. Can u plz type up the working out?
That'll be great!^^
A school invests $5000 at the end of each year at 6% p.a towards a new library. How much will the school have after 10 years?
At the end of year one it has put $5000 in the bank, at an interest rate of 6% to be gained over 9 years.
Amount1: 5000 x (1.06)^9
At the end of the 2nd year, it has put 5000$ in the bank, at an interest rate of 6% to be gained over 8 years (forget about amount 1 at this point)
2nd year: 5000 x (1.06) ^8
as we continue ...we'll eventually get
9th year: 5000 x (1.06) [only one year of interest left]
10th year: 5000 [no interest gained since the amount has been put at the
end of the year]
so we now have 10 seperate amounts.
what we want is the total.
if we add all the amounts together we'll get:
[5000(1.06)^9] + [5000(1.06)^8] ...... + 5000
5000 is a common factor so we can take it out.
= 5000 ( 1.06^9 + 1.06^8 + 1.06^7 .....1)
if you forget this 5000...the bit in the brackets is actually a geometric series.
we can get its total sum.
at this stage what i do is forget about the last term for a while...ie the one with no interest, becoz essentially the 10th year = 5000 + (1.06) ^ 0
and in a geometric series when you add them all up using the Sn formula, its easier to start with the first term not the zeroth term.
Sum of 9 terms = [1.06 (1.06^9 - 1) ]/(1.06 - 1) using formula
suppose the answer to that is X (i cbb using the calculator - sorry
)
so the sum of the 9 terms = X.
multiply that by the 5000 forgot about a little while back
so: 5000X
now since you also forgot about the 10th term, add that back on, so you get
5000X + 5000
should work
p.s bits in bold is the working you're meant to show, all the rest was just for explanation sake
hth
