Answer
$\left(\begin{array}{l} 20\\ 0 \end{array}\right)x^{20}(2y)^0=x^{20}$
$\left(\begin{array}{l} 20\\ 1 \end{array}\right)x^{19}(2y)^1=40x^{19}y$
$\left(\begin{array}{l} 20\\ 2\end{array}\right)x^{18}(2y)^{2}=760x^{18}y^{2}$
Work Step by Step
We find the first 3 terms in the expansion of $(x+2y)^{20}$ by using Binomial Theorem coefficients:
$\left(\begin{array}{l} 20\\ 0 \end{array}\right)x^{20}(2y)^0=1*x^{20}(2y)^0=x^{20}$
$\left(\begin{array}{l} 20\\ 1 \end{array}\right)x^{19}(2y)^1=20*2x^{19}y=40x^{19}y$
$\left(\begin{array}{l} 20\\ 2\end{array}\right)x^{18}(2y)^{2}=190*4x^{18}y^2=760x^{18}y^{2}$