Question Number 22491 by Tinkutara last updated on 19/Oct/17
$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{r}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\:−\:\mathrm{2}{x}\right)^{−\mathrm{1}/\mathrm{2}} \:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\left({r}!\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\mathrm{2}^{{r}} \:\left({r}!\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\left({r}!\right)^{\mathrm{2}} \:\mathrm{2}^{\mathrm{2}{r}} } \\ $$$$\left(\mathrm{4}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\mathrm{2}^{{r}} \:\left({r}\:+\:\mathrm{1}\right)!\left({r}\:−\:\mathrm{1}\right)!} \\ $$
Answered by ajfour last updated on 19/Oct/17
![coeff. of x^r in the expansion of (1+x)^n =((n(n−1)(n−2)..(n−r+1))/(r!)) =(1/2^r )×((2n(2n−2)(2n−4)..(2n−2r+2))/(r!)) coeff. of x^r in the expansion of (1−2x)^(−1/2) is then = (1/2^r )×(((−1)^r [1×3×5×7×...×(2r−1)])/(r!))(−2)^r =((1×3×5×7×....×(2r−1))/(r!)) = ((1×2×3×4×5×6×7×8×...(2r−1)×(2r))/((2)^r ×r!×r!)) =(((2r)!)/((2^r )(r!)^2 )) . option (2) .](Q22496.png)
$${coeff}.\:{of}\:{x}^{{r}} \:{in}\:{the}\:{expansion}\:{of} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} =\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)..\left({n}−{r}+\mathrm{1}\right)}{{r}!} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}^{{r}} }×\frac{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{2}\right)\left(\mathrm{2}{n}−\mathrm{4}\right)..\left(\mathrm{2}{n}−\mathrm{2}{r}+\mathrm{2}\right)}{{r}!} \\ $$$${coeff}.\:{of}\:{x}^{{r}} \:{in}\:{the}\:{expansion}\:{of} \\ $$$$\left(\mathrm{1}−\mathrm{2}{x}\right)^{−\mathrm{1}/\mathrm{2}} \:\:\:\:{is}\:{then} \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{r}} }×\frac{\left(−\mathrm{1}\right)^{{r}} \left[\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}×...×\left(\mathrm{2}{r}−\mathrm{1}\right)\right]}{{r}!}\left(−\mathrm{2}\right)^{{r}} \\ $$$$=\frac{\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}×....×\left(\mathrm{2}{r}−\mathrm{1}\right)}{{r}!} \\ $$$$=\:\frac{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}×\mathrm{6}×\mathrm{7}×\mathrm{8}×...\left(\mathrm{2}{r}−\mathrm{1}\right)×\left(\mathrm{2}{r}\right)}{\left(\mathrm{2}\right)^{{r}} ×{r}!×{r}!} \\ $$$$\:\:\:\:\:=\frac{\left(\mathrm{2}{r}\right)!}{\left(\mathrm{2}^{{r}} \right)\left({r}!\right)^{\mathrm{2}} }\:.\:\:\:\:\:\:\:\:\:\boldsymbol{{option}}\:\left(\mathrm{2}\right)\:. \\ $$
Commented by Tinkutara last updated on 19/Oct/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$