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anonymous
 one year ago
\[a_{n+2}(2^{n+1}+2^{n}) a_{n+1}+a_n=0\]\(a_1=1, a_2=3\).
Find \(a_n\).
anonymous
 one year ago
\[a_{n+2}(2^{n+1}+2^{n}) a_{n+1}+a_n=0\]\(a_1=1, a_2=3\). Find \(a_n\).

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What have you tried?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I saw this problem before and know a solution which is somewhat complicated. I'm going to wait for other solutions :) I think this problem is composed by mukushla, he wants us to try...

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Here's my thoughts so far: The ordinary power series generating function satisfies this equation: \[f(x)\frac{2}{x} f(\frac{x}{2})2f(2x)\frac{1}{x}=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Empty you study physics? That seems like a physicist approach!

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I study everything I can get my hands on lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0good :) btw, how you got that?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Sorry that wasn't quite the link I wanted, this is the book I'm reading currently generatingfunctionology: https://www.math.upenn.edu/~wilf/DownldGF.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{split} a_n &= \left(2^{n+1}+2^{n}\right)a_{n1}a_{n2} =ba_{n1}a_{n2}\\ &= b\left(ba_{n2}a_{n3}\right)a_{n2} \\ &= \left(b^21\right)a_{n2}a_{n3} \\ &= \left(b^3b1\right)a_{n3}a_{n4} \\ &= \left(b^{k+1}\sum_{m=0}^{k1}b^m\right)a_{nk1}a_{nk2} \\ &= \left(b^{k+1}\sum_{m=0}^{k1}b^m\right)a_{nk1}a_{nk2} \\ &= \left(b^{k+1}\frac{1b^{k}}{1b}\right)a_{nk1}a_{nk2} \\ \end{split}\]For our base case, we plug in \(k=n3\): \[ a_n = \left(b^{n2}\frac{1b^{n3}}{1b}\right)a_2+a_1 = 3\left(b^{n2}\frac{1b^{n3}}{1b}\right)+1 \]Then substitute the \(b=2^{n+1}+2^{n}\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It looks good, maybe you have some little mistakes, find \(a_3\) from your formula and once again directly from the relation and compare, see if the formula is right or not :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{split} a_n &= \left(2^{n+1}+2^{n}\right)a_{n1}a_{n2} =ba_{n1}a_{n2}\\ &= b\left(ba_{n2}a_{n3}\right)a_{n2} \\ &= \left(b^21\right)a_{n2}a_{n3} \\ &= \left(b^3b1\right)a_{n3}a_{n4} \\ &= \left(b^{k}\sum_{m=0}^{k2}b^m\right)a_{nk}a_{nk1} \\ &= \left(b^{k}\sum_{m=0}^{k2}b^m\right)a_{nk}a_{nk1} \\ &= \left(b^{k}\frac{1b^{k1}}{1b}\right)a_{nk}a_{nk1} \\ \end{split}\]For our base case, we plug in \(k=n2\): \[ a_n = \left(b^{n2}\frac{1b^{n3}}{1b}\right)a_2+a_1 = 3\left(b^{n2}\frac{1b^{n3}}{1b}\right)+1 \]Then substitute the \(b=2^{n+1}+2^{n}\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There is a special case at \(n=3\) because for that case, we have the coefficient \(b\) rather than \(b\frac{1}{1b}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In fact, all the \(n<4\) cases probably will not hold, because that geometric sum assumes higher values for \(n\).
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