Weird Recursion-2

Let , ${ b }_{ n }=\frac { { a }_{ n } }{ { a }_{ n-1 } }$

Then the given recurrence relation becomes, ${ a }_{ n }-8{ a }_{ n-1 }-7{ a }_{ n-2 }=0 .$

The characteristic equation for this linear second order recurrence relation is,

${ x }^{ 2 }-8x-7=0$

$\Rightarrow x = 1 , 7$ .

so ,

${ a }_{ n }=A+B.{ 7 }^{ n }$

putting n=0 and n=1 and then solving for A and B we get,

$A=\frac { 7{ a }_{ 0 }-{ a }_{ 1 } }{ 6 }$

$B=\frac { { a }_{ 1 }-{ a }_{ 0 } }{ 6 }$

Therefore,

${ a }_{ n }=\frac { 7{ a }_{ 0 }-{ a }_{ 1 } }{ 6 } +\frac { { a }_{ 1 }-{ a }_{ 0 } }{ 6 } .{ 7 }^{ n }$ ------ ( i )

Now,

${ b }_{ 1 }=\frac { { a }_{ 1 } }{ { a }_{ 0 } } \\ \Rightarrow { a }_{ 1 }=b_{ 1 }.{ a }_{ 0 }$

so equation (i) becomes,

${ a }_{ n }=\frac { { a }_{ 0 } }{ 6 } \left\{ 7-b_{ 1 }+(b_{ 1 }-1).{ 7 }^{ n } \right\}$ ------ (ii)

replace $n$ by $n-1$ ,

${ a }_{ n-1 }=\frac { { a }_{ 0 } }{ 6 } \left\{ 7-b_{ 1 }+(b_{ 1 }-1).{ 7 }^{ n-1 } \right\}$ ------- (iii)

calculating $b_{ 1 }=6$

and ,

$\frac { (ii) }{ (iii) } \Rightarrow \frac { { a }_{ n } }{ a_{ n-1 } } =b_{ n }=\frac { 1+5\cdot { 7 }^{ n } }{ 1+5\cdot { 7 }^{ n-1 } }$

as question format A+B+C= 1 + 5 + 7 = 13

and , $\lim _{ n\rightarrow \infty }{ b_{ n } } = 7 = F$

finally $A+B+C+F=13+7=\boxed { 20 }$