Solve Complex Function: Step-by-Step Guide

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The discussion focuses on solving the equation f(n) = 15^n + 8^(n-2) and specifically on deriving f(n+1) - 8f(n) as a multiple of 15^n. The user initially struggled with the first two steps of the solution but later provided the equation for f(n+1) and the calculation for 8f(n). Upon simplifying, the terms involving 8^(n-1) cancel out, leading to the expression (15 - 8)15^n. The key takeaway is the importance of careful manipulation of terms to isolate the desired multiple of 15^n. Understanding these steps is crucial for tackling similar problems in complex functions.
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Hi,
please find attached the problem and the answer. I can't anderstand the first and second step of the answer. I only need a link for the topic or maybe method the used. Bdw I haven't tried to figure out the second answer but for now I only need help tackling the first one.
 

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thanks for telling.. I uploaded it now :)
 
I have the feeling I have done this before!

You have the equation f(n)= 15^n+ 8^{n-2} and want to write f(n+1)- 8 f(n) as a multiple of 15^n.

Okay, f(n+1)= 15^{n+1}+ 8^{n+1- 2}= 15^{n+1}- 8^{n-1} and 8f(n)= 8(15^n)- 8^{n-2})=8(15^n)- 8(8^{n-2})=8(15^n)- 8^{n-1}

Now, when you subtract the two "8^{n-1}" terms cancel and you get 15^{n+1}- 8(15^n)= 15(15^n)- 8(15^n)= (15- 8)15^{n+1}.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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