7. Let $\left(1+x-x^{2}\right)^{10}=a_{0}+a_{1} x+\cdots+a_{20} x^{20}$. Then $a_{0}+a_{1}+2 a_{2}+\cdots+20 a_{20}=$ $\qquad$
7. -9 .

Let $x=0$, we get $a_{0}=1$.
Differentiating both sides of the given equation, we get
$$
\begin{array}{l}
10\left(1+x-x^{2}\right)^{9}(1-2 x) \\
=a_{1}+2 a_{2} x+\cdots+20 a_{20} x^{19} . \\
\text { Let } x=1 \text {, we get. } \\
a_{1}+2 a_{2}+\cdots+20 a_{20}=-10 .
\end{array}
$$

Then $a_{0}+a_{1}+2 a_{2}+\cdots+20 a_{20}=-9$.