Coefficient of $x^7$ in binomial expansion of $(1/6-3x)^{17}$

Question

How would I determine the coefficient of $x^7$ in the expansion of $(1/6-3x)^{17}$ and show the answer as a fraction?

Answer 1

$$(1/6-3x)^{17}=\sum_{k=0}^{17}\binom{17}{k}(1/6)^{17-k}(-3x)^k$$ for $k=7$ we have $$\binom{17}{7}(1/6)^{17-7}(-3x)^7$$ coefficient is $$-\binom{17}{7}\frac{3^7}{6^{10}}=-\binom{17}{7}\frac{1}{3^3\cdot2^{10}}$$

Answer 2

For the coefficient of $x^k$ in (a + bx)^n, we have
$n \choose k$$a^{n-k}b^k$

Putting in n=17 and and k=7, we have,
${17 \choose 7}(1/6)^{10}(-3)^{7} =19448*(1/60466176)*(-2187)=\boxed{-0.703414352}$

Answer 3

From this,

the coefficient of of $x^r,$ in $(a+bx)^n$ is $$\binom nr a^{n-r}b^r$$ where integer $r\in[0,n]$