We have, ${\left(1+x+x^2\right)}^{20}=\sum _{r=0}^{40}{a}_r{x}^r$

$x$ is replaced by $-\frac{1}{x}$, we get

${\left(1-\frac{1}{x}+\frac{1}{x^2}\right)}^{20}=\sum _{r=0}^{40}(-1{)}^r{a}_r{\left(\frac{1}{x}\right)}^r$

${\left(1-x+x^2\right)}^{20}=\sum _{r=0}^{40}(-1{)}^r{a}_r{x}^{40-r}$

$\sum _{r=0}^{39}{\left(-1\right)}^r{a}_r·a_{r+1}$ is coefficient of $x$ in product of ${\left(1+x+x^2\right)}^{20}{\left(1-\frac{1}{x}+\frac{1}{x^2}\right)}^{20}$

Now, ${\left(1+x+x^2\right)}^{20}{\left(1-\frac{1}{x}+\frac{1}{x^2}\right)}^{20}={\left(1+x^2+\frac{1}{x^2}\right)}^{20}$

Here, there will be no term of $x$.

Hence, the coefficient of $x$ will be $0$.