consider the binomial series.\n1 / √(9 - x)=1 / 3(1 - x / 9)^(-1/2)=1 / 3 ∑(n = 0 to ∞)(-1/2 n)( )^n\n=1 /…

Explanation:

Step1: Recall binomial series formula

The binomial series for $(1 + y)^k=\sum_{n = 0}^{\infty}\binom{k}{n}y^{n}=1+ky+\frac{k(k - 1)}{2!}y^{2}+\frac{k(k - 1)(k - 2)}{3!}y^{3}+\cdots+\frac{k(k - 1)\cdots(k - n+1)}{n!}y^{n}+\cdots$, where $\binom{k}{n}=\frac{k(k - 1)\cdots(k - n + 1)}{n!}$ and $|y|\lt1$. In our case, $y=-\frac{x}{9}$ and $k =-\frac{1}{2}$.

Step2: Fill in the first - blank

For the binomial series expansion $\sum_{n = 0}^{\infty}\binom{k}{n}y^{n}$, when $y =-\frac{x}{9}$ and $k=-\frac{1}{2}$, the general term is $\binom{-\frac{1}{2}}{n}\left(-\frac{x}{9}\right)^{n}$, so the first - blank is $-\frac{x}{9}$.

Step3: Fill in the second - blank

The second - term in the binomial expansion $(1 + y)^k$ for $k =-\frac{1}{2}$ and $y=-\frac{x}{9}$ is $\frac{k(k - 1)}{2!}y^{2}$. Substituting $k =-\frac{1}{2}$ and $y =-\frac{x}{9}$, we have $\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2!}\left(-\frac{x}{9}\right)^{2}$, so the second - blank is $-\frac{1}{2}-1=-\frac{3}{2}$.

Step4: Fill in the third - blank

The coefficient of $x$ in the expansion of $\frac{1}{3}\left[1+\left(-\frac{1}{2}\right)\left(-\frac{x}{9}\right)+\cdots\right]$ is $\frac{1}{3}\times\left(-\frac{1}{2}\right)\times(- 1)\times\frac{1}{9}=\frac{1}{54}$.

Answer:

First blank: $-\frac{x}{9}$; Second blank: $-\frac{3}{2}$; Third blank: $\frac{1}{54}$