Let we have a binomial expression \[{{(1+{{x}^{2}})}^{2}}{{(1+x)}^{n}}={{A}_{0}}+{{A}_{1}}x+{{A}_{2}}{{x}^{2}}.......\] If ${{A}_{0}},{{A}_{1}},{{A}_{2}}$ ​ are in A.P, then the value of n

Hint: We will find the value of ${{A}_{0}},{{A}_{1}},{{A}_{2}}$ as we know the relation between them. We will calculate the value of $A_0$ by putting $x=0$ in the given equation. We will now differentiate the given equation w.r.t x and put $x=0$, we will get the value of $A_1$. We will double differentiate the equation w.r.t x and put $x=0$ and we will get the value of $A_2$. We know that if three numbers a, b, c are in A.P then $2b= a+c$ we will apply this relation to get the value of n.

Complete step-by-step solution:
We have the equation \[{{(1+{{x}^{2}})}^{2}}{{(1+x)}^{n}}={{A}_{0}}+{{A}_{1}}x+{{A}_{2}}{{x}^{2}}.......\]
We will now put x=0 in the above equation, we will get,
$\begin{align}
  & \Rightarrow {{(1+{{x}^{2}})}^{2}}{{(1+x)}^{n}}={{A}_{0}}+{{A}_{1}}x+{{A}_{2}}{{x}^{2}}....... \\
 & \Rightarrow {{\left( 1+0 \right)}^{2}}{{\left( 1+0 \right)}^{n}}={{A}_{0}} \\
 & \Rightarrow {{A}_{0}}=1 \\
\end{align}$
We will now differentiate the given equation w.r.t x
 $\begin{align}
  & \Rightarrow {{(1+{{x}^{2}})}^{2}}{{(1+x)}^{n}}={{A}_{0}}+{{A}_{1}}x+{{A}_{2}}{{x}^{2}}....... \\
 & \Rightarrow \dfrac{d{{(1+{{x}^{2}})}^{2}}{{(1+x)}^{n}}}{dx}=0+{{A}_{1}}+2{{A}_{2}}x \\
\end{align}$
We apply product rule in LHS, which is given by
$\begin{align}
  & \Rightarrow \dfrac{d(uv)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx} \\
 & \Rightarrow \dfrac{d{{(1+{{x}^{2}})}^{2}}{{(1+x)}^{n}}}{dx}={{(1+{{x}^{2}})}^{2}}\dfrac{d{{(1+x)}^{n}}}{dx}+{{(1+x)}^{n}}\dfrac{d{{(1+{{x}^{2}})}^{2}}}{dx} \\
 & \Rightarrow {{(1+{{x}^{2}})}^{2}}\times n{{(1+x)}^{n-1}}+{{(1+x)}^{n}}\times 2(1+{{x}^{2}})\times 2x \\
 & \Rightarrow {{(1+{{x}^{2}})}^{2}}\times n{{(1+x)}^{n-1}}+{{(1+x)}^{n}}\times 2(1+{{x}^{2}})\times 2x={{A}_{1}}+2x{{A}_{2}}...........(i) \\
\end{align}$
We will put x=0, we will get,
$\begin{align}
  & \Rightarrow {{(1+{{0}^{2}})}^{2}}\times n{{(1+0)}^{n-1}}+{{(1+0)}^{n}}\times 2(1+{{0}^{2}})\times 2\times 0={{A}_{1}}+2\times 0\times {{A}_{2}} \\
 & \Rightarrow {{A}_{1}}=n \\
\end{align}$
We will now differentiate equation(i) w.r.t x, we will get
$\Rightarrow \dfrac{d}{dx}\left( {{(1+{{x}^{2}})}^{2}}\times n{{(1+x)}^{n-1}} \right)+\dfrac{d}{dx}\left( {{(1+x)}^{n}}\times 2(1+{{x}^{2}})\times 2x \right)=0+2{{A}_{2}}$
We apply product rule in LHS, which is given by
\[\begin{align}
  & \Rightarrow \dfrac{d(uv)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx} \\
 & \Rightarrow \dfrac{d}{dx}\left( {{(1+{{x}^{2}})}^{2}}\times n{{(1+x)}^{n-1}} \right)+\dfrac{d}{dx}\left( {{(1+x)}^{n}}\times 2(1+{{x}^{2}})\times 2x \right) \\
 & \Rightarrow {{(1+{{x}^{2}})}^{2}}\dfrac{d}{dx}\left( n{{(1+x)}^{n-1}} \right)+n{{(1+x)}^{n-1}}\dfrac{d}{dx}\left( {{(1+{{x}^{2}})}^{2}} \right)+{{(1+x)}^{n}}\dfrac{d}{dx}\left( 4x(1+{{x}^{2}}) \right)+4x(1+{{x}^{2}})\dfrac{d}{dx}\left( {{(1+x)}^{n}} \right) \\
 & \Rightarrow {{(1+{{x}^{2}})}^{2}}\times n(n-1){{(1+x)}^{n-2}}+n{{(1+x)}^{n-1}}\times 4x(1+{{x}^{2}})+{{(1+x)}^{n}}\times 4(1+3{{x}^{2}})+4x(1+{{x}^{2}})\times n{{(1+x)}^{n-1}} \\
\end{align}\]
We will now put x=0, we will get
$\begin{align}
  & \Rightarrow n\left( n-1 \right)+4 \\
 & \Rightarrow 2{{A}_{2}}=n\left( n-1 \right)+4 \\
 & \Rightarrow {{A}_{2}}=\dfrac{n\left( n-1 \right)+4}{2} \\
\end{align}$
We know that ${{A}_{0}},{{A}_{1}},{{A}_{2}}$ are in A.P
$\begin{align}
  & \Rightarrow 2{{A}_{1}}={{A}_{0}}+{{A}_{2}} \\
 & \Rightarrow 2n=1+\dfrac{n(n-1)+4}{2} \\
 & \Rightarrow 4n=2+{{n}^{2}}-n+4 \\
 & \Rightarrow {{n}^{2}}-5n+6=0 \\
\end{align}$
We will now solve the above equation to get the value of x
$\begin{align}
  & \Rightarrow {{n}^{2}}-5n+6=0 \\
 & \Rightarrow {{n}^{2}}-3n-2n+6=0 \\
 & \Rightarrow n(n-3)-2(n-3)=0 \\
 & \Rightarrow (n-2)(n-3)=0 \\
 & \Rightarrow n=2,3 \\
\end{align}$
So, The value of n can be 2 and 3.

Note: Caution should be taken to avoid mistakes while solving the differentiation as it gets confusing at times with many terms in it. While solving differentiation the signs should be kept in mind. No, need to simplify the LHS side as it will take a lot of time and we only require the value of ${{A}_{0}},{{A}_{1}},{{A}_{2}}$ which we will get by putting $x=0$, so of the terms will get eliminated. If the three numbers a, b, c are in A.P then the common difference between the numbers is the same which means $b-a = c-b$.