agpanora19
agpanora19
20-03-2017
Mathematics
contestada
an=an-1+6an-2
n>2 a0=3 a1=6
Respuesta :
LammettHash
LammettHash
27-03-2017
[tex]\begin{cases}a_0=3\\a_1=6\\a_n=a_{n-1}+6a_{n-2}&\text{for }n>2\end{cases}[/tex]
Let the generating function for [tex]a_n[/tex] be
[tex]A(x)=\displaystyle\sum_{n\ge0}a_nx^n[/tex]
Multiplying both sides by [tex]x^{n-2}[/tex]
[tex]a_nx^{n-2}=a_{n-1}x^{n-2}+6a_{n-2}x^{n-2}[/tex]
Summing both sides over the non-negative integers greater than or equal to 2 gives
[tex]\displaystyle\sum_{n\ge2}a_nx^{n-2}=\sum_{n\ge2}a_{n-1}x^{n-2}+6\sum_{n\ge2}a_{n-2}x^{n-2}[/tex]
[tex]\displaystyle\frac1{x^2}\sum_{n\ge2}a_nx^n=\frac1x\sum_{n\ge1}a_nx^n+6\sum_{n\ge0}a_nx^n[/tex]
[tex]\displaystyle\frac1{x^2}\left(\sum_{n\ge0}a_nx^n-a_0-a_1x\right)=\frac1x\left(\sum_{n\ge0}a_nx^n-a_0\right)+6\sum_{n\ge0}a_nx^n[/tex]
[tex]\displaystyle\frac1{x^2}\left(A(x)-3-6x\right)=\frac1x\left(A(x)-3\right)+6A(x)[/tex]
[tex]A(x)=\dfrac{3x+3}{1-x-6x^2}[/tex]
[tex]A(x)=\dfrac3{5(1+2x)}+\dfrac{12}{5(1-3x)}[/tex]
For [tex]|x|<\dfrac13[/tex], the two series converge to
[tex]\displaystyle A(x)=\frac35\sum_{n\ge0}(-2x)^n+\frac{12}5\sum_{n\ge0}(3x)^n[/tex]
[tex]\implies A(x)=\displaystyle\sum_{n\ge0}\dfrac{3(-2)^n+12(3)^n}5x^n[/tex]
[tex]a_n=\dfrac{3(-2)^n+12(3)^n}5=\dfrac{3(-2)^n+4(3)^{n+1}}5[/tex]
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