Question

An=an-1+6an-2
n>2 a0=3 a1=6

Answer 1

`\begin{cases}a_0=3\\a_1=6\\a_n=a_(n-1)+6a_(n-2)&\text{for }n>2\end{cases}`

Let the generating function for
`a_n` be


`A(x)=\displaystyle\sum_(n\ge0)a_nx^n`

Multiplying both sides by
`x^(n-2)`


`a_nx^(n-2)=a_(n-1)x^(n-2)+6a_(n-2)x^(n-2)`

Summing both sides over the non-negative integers greater than or equal to 2 gives


`\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)`

`\displaystyle\frac1{x^2}\sum_(n\ge2)a_nx^n=\frac1x\sum_(n\ge1)a_nx^n+6\sum_(n\ge0)a_nx^n`

`\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`

`\displaystyle\frac1{x^2}\left(A(x)-3-6x\right)=\frac1x\left(A(x)-3\right)+6A(x)`

`A(x)=(3x+3)/(1-x-6x^2)`

`A(x)=\frac3{5(1+2x)}+(12)/(5(1-3x))`

For
`|x|<\frac13`, the two series converge to


`\displaystyle A(x)=\frac35\sum_(n\ge0)(-2x)^n+\frac{12}5\sum_(n\ge0)(3x)^n`

`\implies A(x)=\displaystyle\sum_(n\ge0)\frac{3(-2)^n+12(3)^n}5x^n`

`a_n=\frac{3(-2)^n+12(3)^n}5=\frac{3(-2)^n+4(3)^(n+1)}5`