Repayments help (1 Viewer)

[nazfiz]

Can someone please help me with part V of the question.

THANKS! in advance

 

[jdnRof]

Okay this will be a bit messy to write up, but here goes my attempt at explaining:

I'm gonna call the amount leftover each time F(n), where n is the number of award ceremonies.

So we know that F(1)= (10,000 x 1.03) - 450 *same as part a)

Since we are going to use an extra 2% each year, then we can say that in the second year, we instead withdraw 450 x 1.02 from the money

F(2) = [(10,000 x 1.03) - 450] x 1.03 - 450 x 1.02
= 10,000 x 1.03² - 450 (1.03 + 1.02)

In the third year, once again, we withdraw an extra 2%, so that would be 450 x 1.02²

F(3) = {10,000 x 1.03² - 450 (1.03 + 1.02)} x 1.03 - 450 x 1.02²
= 10,000 x 1.03³ -450 (1.03² + 1.02 x 1.03 + 1.02²)

From here we can generalise:

F(n) = 10,000 x 1.03ⁿ - 450 (1.03^n-1 + 1.03^n-2 x 1.02 + .... + 1.03 x 1.02^n-2 + 1.03 x 1.02^n-1)


I assume you can derive the formula from there with a bit of simplifying and using GP setting the ratio to 1.02/1.03