8. What is the coefficient for $x^{5}$ in the expansion $(x-2)^{7}$ ?
Answer
Thus, the coefficient for $x^{5}$ in the expansion $(x-2)^{7}$ is \(\boxed{84}\).
Step 1 ：Given the expression $(x-2)^{7}$, we want to find the coefficient of $x^{5}$.
Step 2 ：Using the binomial theorem, which states that $(a+b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}$, we can identify $a=x$, $b=-2$, and $n=7$.
Step 3 ：We need to find the term where $n-k=5$, which means $k=2$.
Step 4 ：Therefore, the coefficient of $x^{5}$ is calculated as $\binom{7}{2} (-2)^{2}$.
Step 5 ：Calculating the binomial coefficient $\binom{7}{2}$ gives us 21.
Step 6 ：Multiplying this by $(-2)^{2}$ gives us 84.
Step 7 ：Thus, the coefficient for $x^{5}$ in the expansion $(x-2)^{7}$ is \(\boxed{84}\).
