Applications of a GP Help :) (1 Viewer)

[Veritude]

Hey guys I've been stuck on this question and would appreciate someone to guide me through how to do it

A fund is established to provide prizes for a cricket team's annual Awards night. $10,000 is place in the fund one year before the first Awards night. It is decided that $450 will be withdrawn from the fund each year to purchase the annual prizes. The money in the fund is invested at 3% p.a. compounded annually with the interest paid into the fund before each annual Awards night.

v) For the fund described above, it is decided to increase the amount of money withdrawn for each Awards night by 2% each year.

Show that An = 45,000(1.02)^n - 35,000(1.03)^n.

I'm not sure where to put this question as it's my first post and I don't even know whether its 3u/2u but yeah it would be fantastic if someone showed me how to do it.

 

[asianese]

Could you please quote the whole question? It would help.

 

[Veritude]

Sure,

i) Show that the fund contains $9695.50 after the second awards night.
ii) If An is the amount in dollars remaining in the fund after the nth Awards night, prove that An = 5000(3-1.03^n)
iii) Find the amount of money in the fund after the 25th Awards night.
iv) Find the maximum number of Awards nights that can be financed using this fund.
v) I've already written this.

Thanks a lot for your help!

 

[Shadowdude]

Okay, got it!

Very good question. Had me tempted to go into binomial coefficients for a bit until I realised... "oh wait".




Then we do a bit of 'play' (as one of my lecturers calls it), just to find a pattern:







So:



At this point I see a pattern, and claim:



Convince yourself that is true.

Hard bit is seeing that this is indeed a geometric progression. But the common ratio is:




I originally thought: 'Hmm, is in form a^n + a^(n-1)b + ... + b^n , so something to do with binomials'.

Anyways - so for the general case, it becomes:



There are n-1 terms in this progression, use your formula:







Now use a 'trick' so you get ___(1.03)^{n}

 

[Veritude]

Okay, got it!

Thank you so much for your help Shadowdude I really appreciate it I can guess this took a while to do! I understand most of it until it goes up to the part which I quoted, would you mind explaining how you arrived to these conclusions (from the first part to second and then second to third)? Thanks a lot!

 

[Shadowdude]

Thank you so much for your help Shadowdude I really appreciate it I can guess this took a while to do! I understand most of it until it goes up to the part which I quoted, would you mind explaining how you arrived to these conclusions (from the first part to second and then second to third)? Thanks a lot!

An hour or so. About half an hour last night, and then another half hour this morning... mostly just thinking about it. But yeah, once you spot the trick it's always like "OHHHHHHHHHHHH..."

Gah, haven't done any of these in like two years.


To your queries:

From 1 to 2: I expanded the brackets, basically. I take the denominator out and note that 1 - (1.02/1.03) = 0.01/1.03 = 1/103. And then the rest should follow.

From 2 to 3: 450/(1/103) = 46350. Now, for the last summand (I think that's what it's called?), 1.03^(n-1) is on top, 1.03^n is on the bottom, so you simplify that and you get 1/1.03.


I hope that helps.