<s>[INST] 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..
 [/INST] In line 4, $(n+1/2) = (2n+1)/2$ for all n running from 0 to r-1 and in denominator there become 2's r times which give $2^r$ in denominator. In next line they have multiplied $1.2...r$ and they distribute 2's for each n  running from 1 to r which becomes $2.4....2r$ ($2^r$ is in numerator too). In next line they take back 2's which becomes $2^r$ in denominator and in numerator they rearrange each term which become $(2r)!$.
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