If (1 + x + x^ 2 )20 = ∑ _ r = 0 ^ 40 a _ r · x^ r , then ∑ _ r = 0...

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

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

\left(1 - \dfrac{1}{x} + \dfrac{1}{x2}\right)20 = \mathop{∑}_{r = 0}^{40} ( - 1{)}^{r}{a}_{r}\left(\dfrac{1}{x}\right)r$</p><p>\left(1 - x + x^{2}\right)20 = \mathop{∑}_{r = 0}^{40} ( - 1{)}^{r}{a}_{r}x^{40 - r}$

$\mathop{∑}_{r = 0}^{39}\left(- 1\right)r{a}_{r} \cdot {a}_{r + 1}$ is coefficient of $x$ in product of~\left(1 + x + x^{2}\right)20\left(1 - \dfrac{1}{x} + \dfrac{1}{x2}\right)20$</p><p>Now,~\left(1 + x + x^{2}\right)20\left(1 - \dfrac{1}{x} + \dfrac{1}{x2}\right)20 = \left(1 + x^{2} + \dfrac{1}{x2}\right)20$

Here, there will be no term of~$x$.

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