The absolute difference of the coefficients of x 10 and x 7 in the expansion of 2 x 2 + 1 2 x 11 is equal to

The given expansion is ${\left(2x^2+\frac{1}{2x}\right)}^{11}$.

The general term in the binomial expansion of ${\left(x+a\right)}^n$ is given by $T_{r+\mathit{1}}\mathit{=}{}^{n}C_r{x}^{n\mathit{-}r}{a}^r$.

$\Rightarrow T_{r+1}={}^{11}C_r{\left(2x^2\right)}^{11-r}{\left(\frac{1}{2x}\right)}^r$

$={}^{11}C_r\left(2^{11-r-r}\right)x^{22-2r-r}$

Now for coefficient of $x^7$$\Rightarrow 22-3r=7$

$\Rightarrow r=5$

Coefficient of $x^7$ is ${}^{11}C_5\left(2^{11-5\times 2}\right)$

Now for coefficient of $x^{10}\Rightarrow 22-3r=10$

$\Rightarrow r=4$

Coefficient of $x^{10}$ is ${}^{11}C_4\left(2^{11-2\times 4}\right)$

We need the absolute difference, we get it by $\left|{}^{11}C_5\left(2^{11-5\times 2}\right)-{}^{11}C_4\left(2^{11-4\times 2}\right)\right|$.

$=\left|924-2640\right|=1716$

$={12}^3-12$

Hence the absolute difference is $={12}^3-12$.