Proof explanation: finding the coefficient of $(r+1)$th term in the expansion of $\left(1-6x\right)^{-\frac{1}{2}}$? Here is the answer of the math..
$\displaystyle\left(1-6x\right)^{-\frac{1}{2}}$
$\displaystyle=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)....\left(-\frac{1}{2}-r+1_{ }\right)}{r!}\left(-6x\right)^r$
$\displaystyle=\frac{\left(-1\right)^r\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)....\left(r-1+\frac{1}{2}\right)}{r!}\left(-1\right)^r.\:x^r.\:2^r.3^r$
$\displaystyle=\left(-1\right)^{2r}\:\frac{1.3.5.7.....\left(2r-1\right)}{2^r\:.\:r!}\:\:x^r.\:2^r.3^r$
$\displaystyle=\frac{\left\{1.3.5.7.....\left(2r-1\right)\right\}\left(2.4.6....2r\right)}{r!\:\left(2.4.6.....2r\right)}\:\:x^r.\:3^r$
$\displaystyle=\frac{1.2.3.4.....2r}{r!\:2^r\:\left(1.2.3.\:....r\right)}\:\:x^r.\:3^r$
$\displaystyle=\frac{\left(2r\right)!\:\:.\:3^r}{2^r\:.\:\left(r!\right)^2}\:\:x^r$
$\displaystyle=\left(\frac{3}{2}\right)^r\:\frac{\left(2r\right)!\:}{\:\left(r!\right)^2}\:.\:\:x^r$
I can't understand the 4th and 6th line of this math . Please explain me that lines..


We obtain
  \begin{align*}
&\color{blue}{\frac{\left(-1\right)^r\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)\cdots\left(r-1+\frac{1}{2}\right)}{r!}\left(-1\right)^rx^r2^r3^r}\\
&\qquad=\frac{\left(-1\right)^r\left(\frac{1}{2}\right)\left(\frac{1+2}{2}\right)\left(\frac{1+4}{2}\right)\cdots\left(\frac{2r-2+1}{2}\right)}{r!}\left(-1\right)^rx^r2^r3^r\tag{1}\\
&\qquad=\frac{\left(-1\right)^{2r}\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{5}{2}\right)\cdots\left(\frac{2r-1}{2}\right)}{r!}x^r2^r3^r\tag{2}\\
&\qquad\color{blue}{=(-1)^{2r}\frac{1\cdot3\cdot5\cdots(2r-1)}{2^rr!}x^r2^r3^r}\tag{3}\\
&\qquad=\frac{1\cdot3\cdot5\cdots(2r-1)\cdot  2\cdot   4\cdot  6\cdots  (2r)}{r!2\cdot   4\cdot  6\cdots  (2r)}x^r3^r\tag{4}\\
&\qquad=\frac{(2r)!}{r!(2\cdot1)\cdot   (2\cdot 2)\cdot  (2\cdot 3)\cdots  (2r)}x^r3^r\tag{5}\\
&\qquad\color{blue}{=\frac{(2r)!}{r!2^r1\cdot   2\cdot 3\cdots  r}x^r3^r}\tag{6}\\
\end{align*}

Common:


*

*In (1) we use a common denominator $2$ for the factors in the numerator.

*In (2) we do a simplification and collect the factors $(-1)^r$.

*In (3) we observe there are $r$ factors $\frac{1}{2}$  which can be written as $2^r$ in the denominator.

*In (4) we see odd factors $1,3,5,\ldots,2r-1$ in the numerator. Since we wish to use the more compact notation of factorials we expand numerator and denominator with corresponding even factors. We also use that an even number of factors $(-1)$ give $1$.

*In (5) we write the numerator using factorial notation and observe that each even factor in the denominator can be written a $2\cdot k$.

*In (6) we factor out $r$ factors $2$ giving $2^r$.
