(7) Let $\left(1+x-x^{2}\right)^{10}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{20} x^{20}$, then $a_{0}+$ $a_{1}+2 a_{2}+3 a_{3}+\cdots+20 a_{20}=$ $\qquad$
(7) -9 Hint: Let $x=0$, we get $a_{0}=1$. Differentiate both sides
$$
\begin{array}{c}
10\left(1+x-x^{2}\right)^{9}(1-2 x)=a_{1}+2 a_{2} x+3 a_{3} x^{2}+\cdots+20 a_{20} x^{19} . \\
\text { Let } x=1, a_{1}+2 a_{2}+3 a_{3}+\cdots+20 a_{20}=-10 \text {, thus } \\
a_{0}+a_{1}+2 a_{2}+3 a_{3}+\cdots+20 a_{20}=-9 .
\end{array}
$$