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Given this problem with solution:

http://postimg.org/image/gouhieo35/

I have a really simple question that i still can't understand. When he divided by $4^n$ how did 8C = 2C, 16C = C, and he cancelled out $4^{n-2}$ and $4^{n-1}$?

iadvd
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user2294579
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1 Answers1

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No, no. We have \begin{align} &{}\frac{Cn^2 4^n - 8C(n-1)^2 4^{n-1} + 16C(n-2)^2 4^{n-2}}{4^n} =\\ &= \frac{Cn^2 4^n}{4^n} - \frac{8C(n-1)^2 4^{n-1}}{4^n} + \frac{16C(n-2)^2 4^{n-2}}{4^n}=\\ &=Cn^2 - \frac{8C(n-1)^2}{4} + \frac{16C(n-2)^2}{4^2} =\\&= Cn^2 - 2C(n-1)^2 + C(n-2)^2\end{align}

Michael Galuza
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  • So let's say that 16C = 15C. Am i correct in assuming that it will become 15/16 C? or if its just 1 it will become C/16C = 16C^-1? – user2294579 Jun 10 '15 at 05:58
  • @user2294579, I'm not sure that I understand you. Simplification of $Cn^2-2C(n-1)^2 + C(n-2)^2$ is $n^2(C-2C+C) + n[(-2)\cdot(-2)-4]+(-2\cdot 1+4)=2C$, and it's equal to 3, and $C=3/2$. – Michael Galuza Jun 10 '15 at 06:09
  • Oh what im asking is what if its not 16C as shown in the 2nd solution that you wrote and as described by the problem above but is 15C would it then become 15C/4^2 = 15/16C when you divide it by 4^2? or if its just originally 1 will it become C/16C = 16C^-1? – user2294579 Jun 10 '15 at 06:39
  • @user2294579, if $16C=15C$, then $C=0$; it's not interesting) I'm not sure that I understand you, but $(15C)/4^2$ definitely equal to $(15/16)C$. – Michael Galuza Jun 10 '15 at 06:44