If n is a +ve integer greater than 3 such that ${\left(1+{\mathrm{x}}^2\right)}^2{\left(1+\mathrm{x}\right)}^{\mathrm{n}}=\sum _{\mathrm{K}=0}^{\mathrm{n}+4}{\mathrm{a}}_{\mathrm{K}}{\mathrm{x}}^K$ and ${\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3$ are in AP then maximum value of n is _____

$$\begin{gathered} {\left(1+{\mathrm{x}}^2\right)}^2{(1+\mathrm{x})}^{\mathrm{n}}=\sum _{\mathrm{K}=0}^{\mathrm{n}+4}{\mathrm{a}}_{\mathrm{K}}{\mathrm{x}}^{\mathrm{K}} \\ \Rightarrow {\mathrm{a}}_1={\mathrm{nC}}_1,2+{}^{\mathrm{n}}\mathrm{C}_2={\mathrm{a}}_2 \\ and {}^{\mathrm{n}}\mathrm{C}_3+2.{}^{\mathrm{n}}\mathrm{C}_1={\mathrm{a}}_3 \\ \mathrm{But} {\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3 \mathrm{are} \mathrm{in} \mathrm{AP} \\ \Rightarrow 2\left(2{+}^{\mathrm{n}}{\mathrm{C}}_2\right){=}^{\mathrm{n}}{\mathrm{C}}_1{+}^{\mathrm{n}}{\mathrm{C}}_3+2{\cdot }^{\mathrm{n}}{\mathrm{C}}_1 \\ \Rightarrow {\mathrm{n}}^3-9{\mathrm{n}}^2+26\mathrm{n}-24=0 \\ \Rightarrow (\mathrm{n}-2)(\mathrm{n}-3)(\mathrm{n}-4)=\Rightarrow \mathrm{n}=2,3,4 \end{gathered}$$

Maximum value of n is 4