Answer
$13,440a^{4}b^{6}$
Work Step by Step
The $k^{th}$ term of $(a+b)^n$ is given by $\left(\begin{array}{cc}n\\k-1
\end{array}\right)a^{n-k+1}b^{k-1}.$ Thus the $7^{th}$ term of $(a-2b)^{10}$ is
\begin{align*}\require{cancel}
&
\left(\begin{array}{cc}10\\7-1
\end{array}\right)a^{10-7+1}(-2b)^{7-1}
\\\\&=
\left(\begin{array}{cc}10\\6
\end{array}\right)a^{4}(-2b)^{6}
\\\\&=
\left(\begin{array}{cc}10\\6
\end{array}\right)a^{4}(64b^{6})
\\\\&=
64\left(\begin{array}{cc}10\\6
\end{array}\right)a^{4}b^{6}
.\end{align*}
Using $\left(\begin{array}{cc}n\\r
\end{array}\right)=\dfrac{n!}{r!(n-r)!}$, the expression above is equivalent to
\begin{align*}\require{cancel}
&
64\cdot\dfrac{10!}{6!(10-6)!}a^{4}b^{6}
\\\\&=
64\cdot\dfrac{10!}{6!4!}a^{4}b^{6}
\\\\&=
64\cdot\dfrac{10(9)(8)(7)(\cancel{6!})}{\cancel{6!}4!}a^{4}b^{6}
\\\\&=
64\cdot\dfrac{10(9)(\cancel8)(7)}{\cancel4(3)(\cancel2)(1)}a^{4}b^{6}
\\\\&=
64\cdot\dfrac{10(\cancelto39)(7)}{\cancelto13}a^{4}b^{6}
\\\\&=
13440a^{4}b^{6}
.\end{align*}
Hence, the $7^{th}$ term of $(a-2b)^{10}$ is $13,440a^{4}b^{6}$.
