$a_n=6n^2+2n+a_{n-1}$
Solution = Homogeneous Solution + Particular Solution .................$(1)$
$Homogeneous\ Solution$,
$a_n=a_{n-1}$
$a_n-a_{n-1}=0$
let, $a_n = x$
$x -1 = 0$
$x = 1$
$Homogeneous \ Solution=d*1^n = d$.......................................$(2)$
$ Particular\ Solution :$
Here, $F(x) = 6n^2 + 2n$ // Quadratic
let us assume, $a_n = (an^2 + bn + c)*n$ // here root of homogeneous solution is 1 so we have to multiply General quadratic solution by n. ........................$(3)$
$6n^2+ 2n = a_n - a_{n-1}$
$={n*(an^2 + bn + c) - (n-1)*(a(n-1)^2 + b(n-1) + c)}$
$= an^3 + bn^2 + cn - ( an^3 - a -3an^2 +3an + bn^2 + b - 2bn +cn - c )$
$= ( an^3 + bn^2 + cn - an^3 + a +3an^2 -3an - bn^2 - b + 2bn - cn + c )$
$= 3an^2 + (2b-3a)n + (a-b+c)$
Apply Principle of Homogeneity,
$3a =6$ $2b -3a= 2$ $a-b+c =0$
$a = 2$ $b = 4$ $c = 2$
Now put values in equation $(3)$,
$Particular\ Solution = n*(2n^2 + 4n +2) = 2n^3 + 4n^2 + 2n$ ..................$(4)$
from equation $(1), (2)$ and $(4)$
Solution of recurrence $= 2n^3 + 4n^2 + 2n +d$
here $a(1) = 8$ //given
by putting $n = 1, d=0$
Final Solution of recurrence $= 2n^3 + 4n^2 + 2n = 2n(n+1)^2$
Value of $a(99) = 2×99×(10)^4 = 198*10^4$
So, $K$ value is $198$