If $\sum _{r=0}^{3n}{a}_r{(x-4)}^r=\sum _{r=0}^{3n}{A}_r{(x-5)}^r$  and  $a_k=1\forall k\ge 2n$ and $\sum _{r=0}^{3n}{d}_r{(x-8)}^r=\sum _{r=0}^{3n}{B}_r{(x-9)}^r$  and  $\sum _{r=0}^{3n}{d}_r{(x-12)}^r=\sum _{r=0}^{3n}{D}_r{(x-13)}^r$ and $d_k=1\forall k\ge 2n.$  then find the value of $\frac{A_{2n}+D_{2n}}{B_{2n}}$  .

$x-4=t+1\Rightarrow x-5=t$

$\therefore \sum _{r=0}^{3n}{a}_r{(1+t)}^r=\sum _{r=0}^{3n}{A}_r{t}^r$

$\begin{array}{l}{a}_0{(1+t)}^0+a_1{(1+t)}^1+\mathrm{.....}+a_{2n}{(1+t)}^{2n}+a_{2n+1}{(1+t)}^{2n+1}+\mathrm{...}{a}_{3n}{(1+t)}^{3n}\\ =A_0{t}^0+A_1{t}^1+\mathrm{.....}+A_{2n}{t}^{2n}+\mathrm{.....}+A_{3n}{t}^{3n}\end{array}$

compare coefficient of $t^{2n}$ on both sides

$\begin{array}{l}{A}_{2n}=a_{2n}.{}^{2n}C_{2n}+a_{2n+1}{}^{2n+1}C_{2n}+\mathrm{....}+a_{3n}{}^{3n}C_{2n}\\ ={}^{2n}C_{2n}+{}^{2n+1}C_{2n}+{}^{2n+2}C_{2n}+\mathrm{....}+{}^{3n}C_{2n}\\ [\because a_{2n}=a_{2n+1}=\mathrm{....}=a_{3n}=1]\\ =({}^{2n+1}C_{2n+1}+{}^{2n+1}C_{2n})+\mathrm{....}+{}^{3n}C_{2n}\end{array}$

$A_{2n}={}^{3n+!}C_{2n+1}$

Similarly $B_{2n}={}^{3n+!}C_{2n+1}$

$\begin{array}{l}{D}_{2n}={}^{3n+!}C_{2n+1}\\ \therefore \frac{A_{2n}+D_{2n}}{B_{2n}}=2\end{array}$