Let ${(1+x+x^2+x^3+x^4)}^n=a_0+a_{1}x+a_2{x}^2+\dots +a_{4n}{x}^{4n},$ then the value of $\frac{a_1+a_6+a_{11}+\dots }{a_1+a_2+a_3+a_4+a_6+a_7+a_8+a_9+a_{11}\dots }$ is equal to ______

Put $x=1$  we get  $5^n=a_0+a_1+a_2+a_3+a_4+a_5+..........$
Put $\text{x}={\text{e}}^{i\frac{4\pi }{3}}\dots \text{put\hspace{0.17em}}x=e^{i\frac{6\pi }{5}}\dots$
After simplification we get 

$a_0+a_5+a_{10}+\dots =a_1+a_6+a_{11}+\dots =a_2+a_7+a_{12}+\dots =a_3+a_8+a_{13}+\dots \dots \mathrm{..}=$ $a_4+a_9+a_{14}+\dots =\frac{5^n}{5}=5^{n-1}$

$\Rightarrow \text{ Required equals to }\frac{5^{\text{n}-1}}{4\times 5^{n-1}}=0.25$