Answer : $a_n=A(-6)^n+B(1)^n$
Reference :
- https://www.youtube.com/watch?v=Pp4PWCPzeQs
- https://www.youtube.com/watch?v=LNOZlxyrkPA
Solution :
$a_n=-5a_{n-1}+6a_{n-2}$
$\therefore a_n+5a_{n-1}-6a_{n-2}=0$
Let $G(x)$ be generating function of sequence $a_0,a_1,a_2,a_3...$
$G(x)$ $=$ $a_0$ + $a_1x$ + $a_2x^2$ + $a_3x^3…$
$5xG(x)$ $=$ $5a_0x$ + $5a_1x^2$ + $5a_2x^3$ + $5a_3x^4…$
$-6x^2G(x)$ $=$ – $6a_0x^2$ – $6a_1x^3$ – $6a_2x^4$ – $6a_3x^5…$
$\therefore (1+5x-6x^2)G(x)$
$=a_0+a_1x + 5a_0x + (a_2+5a_1-6a_0)x^2 + (a_3+5a_2-6a_1)x^3 + ...$
$=a_0+a_1x + 5a_0x +$ $(0)x^2 +$ $(0)x^3 + ...$
$=a_0+(a_1+ 5a_0)x$
$G(x)=\frac{a_0+(a_1+ 5a_0)x}{(1+5x-6x^2)}=\frac{a_0+(a_1+ 5a_0)x}{(1+6x)(1-x)}=\frac{A}{(1+6x)}+\frac {B}{(1-x)}$
By partial fraction $A=\frac{6a_0-(a_1+ 5a_0)}{7}=\frac {a_0-a_1}{7},B=\frac {a_0+(a_1+ 5a_0)}{7}=\frac{6a_0+a_1}{7}$ but lets ignore that.
Expansion of series $(1-kx)^{-1}=\sum_{n=0}^{\infty}k^nx^n$
$\therefore G(x)=A\sum_{n=0}^{\infty}(-6)^nx^n+B\sum_{n=0}^{\infty}(1)^nx^n=\sum_{n=0}^{\infty}(A(-6)^n+B(1)^n)x^n$
$\therefore a_n=A(-6)^n+B(1)^n$
