Infinite geometric series

[Guest]

Hi all,

I have a question that has puzzled me for a long time and hopefully you can help me with.

I was given the following information:

[tex]f1 = \alpha (2f2+f3+f4+f5...)[/tex]
[tex]f3 = (5/6)f2[/tex]
[tex]f4 = (5/6)f3[/tex]
[tex]f[n] = (2/3)* f[n-1], n > 4[/tex]

with the above ending up expressed in terms of f2 and the main question being, how did they get to this value:

[tex]f1 = \alpha (4.916f2)[/tex]

With my simple mind, I was able to solve this problem using knowledge of infinite geometric series as shown below; however, I've been told that whilst the answer is correct, they want a (perhaps more appropriate???) answer to this problem using differential calculus???. I just can't even imagine where to start with that...any help out there....please!!!

My solution (briefly):

if f3 = (5/6)f2,
then f4 = (25/36)f2

and we have that

[tex]2f2 + f3 = (2 + 5/6)f2
= (17/6)f2[/tex]

The rest of the proportions can be solved as an infinite geometric series with first term being f4 and common ratio 2/3, thus the sum of the series is:

[tex]f4/(1-2/3) = 3f4[/tex]

and then it becomes a simple exercise:

[tex]2f2+f3+f4+f5... = (17/6 + 3(25/36))f2
= (59/12)f2
= 4.916f2[/tex]

I would appreciate any help.

[HallsofIvy]

I cannot imagine why, or how, anyone would use "differential Calculus" to solve this!