10. Let $(1-2 x)^{7}=\sum_{k=0}^{7} a_{k} x^{k}$. Then $2 a_{2}+3 a_{3}+4 a_{4}+5 a_{5}+6 a_{6}+7 a_{7}=$ $\qquad$
10. 0 .

For $(1-2 x)^{7}=\sum_{k=0}^{7} a_{k} x^{k}$, differentiating both sides with respect to $x$ yields
$$
\begin{aligned}
- & 14(1-2 x)^{6} \\
= & a_{1}+2 a_{2} x+3 a_{3} x^{2}+4 a_{4} x^{3}+5 a_{5} x^{4}+ \\
& 6 a_{6} x^{5}+7 a_{7} x^{6} .
\end{aligned}
$$

Substituting $x=0$ and $x=1$ into the above equation, we get
$$
\begin{array}{l}
a_{1}=-14, \\
a_{1}+2 a_{2}+3 a_{3}+4 a_{4}+5 a_{5}+6 a_{6}+7 a_{7}=-14 . \\
\text { Therefore, } 2 a_{2}+3 a_{3}+4 a_{4}+5 a_{5}+6 a_{6}+7 a_{7} \\
=-14-(-14)=0 \text {. }
\end{array}
$$