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Question Number 63937 by gunawan last updated on 11/Jul/19
The coefficient of x^r in the expansion of (1−4x)^(−1/2) is
$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{r}} \:\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}−\mathrm{4}{x}\right)^{−\mathrm{1}/\mathrm{2}} \:\:\mathrm{is} \\ $$
Answered by mr W last updated on 12/Jul/19
C_r ^(1/2) =(1/(r!))×(1/2)×((1/2)−1)×((1/2)−2)×...×((1/2)−(r−1)) C_r ^(1/2) =(1/(r!))×(1/2)×(((1−2)/2))×(((1−2×2)/2))×...×(((1−2(r−1))/2)) C_r ^(1/2) =(((−1)^(r−1) ×1×3×5...×(2r−3))/(2^r r!)) C_r ^(1/2) =(((−1)^(r−1) (2r−3)!!)/(2^r r!)) (√(1−x))=(1−x)^(1/2) =Σ_(r=0) ^∞ C_r ^(1/2) (−x)^r (√(1−x))=−Σ_(r=0) ^∞ (((2r−3)!!)/(2^r r!))x^r (√(1−4x))=−Σ_(r=0) ^∞ ((2^r (2r−3)!!)/(r!))x^r ((d...)/dx) on both sides (1/( (√(1−4x))))=Σ_(r=1) ^∞ ((2^(r−1) (2r−3)!!)/((r−1)!))x^(r−1) (1−4x)^(−(1/2)) =(1/( (√(1−4x))))=Σ_(r=0) ^∞ ((2^r (2r−1)!!)/(r!))x^r ⇒coef. of x^r is a_r =((2^r (2r−1)!!)/(r!)) r=0: a_0 =((2^0 ×1)/1)=1 r=1: a_1 =((2^1 ×1)/1)=2 r=2: a_2 =((2^2 ×3×1)/(2×1))=6 r=3: a_3 =((2^3 ×5×3×1)/(3×2×1))=20 r=4: a_4 =((2^4 ×7×5×3×1)/(4×3×2×1))=70 r=5: a_5 =((2^5 ×9×7×5×3×1)/(5×4×3×2×1))=252 ......
$${C}_{{r}} ^{\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{{r}!}×\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)×\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right)×…×\left(\frac{\mathrm{1}}{\mathrm{2}}−\left({r}−\mathrm{1}\right)\right) \\ $$$${C}_{{r}} ^{\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{{r}!}×\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{1}−\mathrm{2}}{\mathrm{2}}\right)×\left(\frac{\mathrm{1}−\mathrm{2}×\mathrm{2}}{\mathrm{2}}\right)×…×\left(\frac{\mathrm{1}−\mathrm{2}\left({r}−\mathrm{1}\right)}{\mathrm{2}}\right) \\ $$$${C}_{{r}} ^{\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\left(−\mathrm{1}\right)^{{r}−\mathrm{1}} ×\mathrm{1}×\mathrm{3}×\mathrm{5}…×\left(\mathrm{2}{r}−\mathrm{3}\right)}{\mathrm{2}^{{r}} {r}!} \\ $$$${C}_{{r}} ^{\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\left(−\mathrm{1}\right)^{{r}−\mathrm{1}} \left(\mathrm{2}{r}−\mathrm{3}\right)!!}{\mathrm{2}^{{r}} {r}!} \\ $$$$\sqrt{\mathrm{1}−{x}}=\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{{r}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(−{x}\right)^{{r}} \\ $$$$\sqrt{\mathrm{1}−{x}}=−\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{r}−\mathrm{3}\right)!!}{\mathrm{2}^{{r}} {r}!}{x}^{{r}} \\ $$$$\sqrt{\mathrm{1}−\mathrm{4}{x}}=−\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{r}} \left(\mathrm{2}{r}−\mathrm{3}\right)!!}{{r}!}{x}^{{r}} \\ $$$$\frac{{d}…}{{dx}}\:{on}\:{both}\:{sides} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}}}=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{r}−\mathrm{1}} \left(\mathrm{2}{r}−\mathrm{3}\right)!!}{\left({r}−\mathrm{1}\right)!}{x}^{{r}−\mathrm{1}} \\ $$$$\left(\mathrm{1}−\mathrm{4}{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}}}=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{r}} \left(\mathrm{2}{r}−\mathrm{1}\right)!!}{{r}!}{x}^{{r}} \\ $$$$\Rightarrow{coef}.\:{of}\:{x}^{{r}} \:{is}\:{a}_{{r}} =\frac{\mathrm{2}^{{r}} \left(\mathrm{2}{r}−\mathrm{1}\right)!!}{{r}!} \\ $$$${r}=\mathrm{0}:\:{a}_{\mathrm{0}} =\frac{\mathrm{2}^{\mathrm{0}} ×\mathrm{1}}{\mathrm{1}}=\mathrm{1} \\ $$$${r}=\mathrm{1}:\:{a}_{\mathrm{1}} =\frac{\mathrm{2}^{\mathrm{1}} ×\mathrm{1}}{\mathrm{1}}=\mathrm{2} \\ $$$${r}=\mathrm{2}:\:{a}_{\mathrm{2}} =\frac{\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{1}}{\mathrm{2}×\mathrm{1}}=\mathrm{6} \\ $$$${r}=\mathrm{3}:\:{a}_{\mathrm{3}} =\frac{\mathrm{2}^{\mathrm{3}} ×\mathrm{5}×\mathrm{3}×\mathrm{1}}{\mathrm{3}×\mathrm{2}×\mathrm{1}}=\mathrm{20} \\ $$$${r}=\mathrm{4}:\:{a}_{\mathrm{4}} =\frac{\mathrm{2}^{\mathrm{4}} ×\mathrm{7}×\mathrm{5}×\mathrm{3}×\mathrm{1}}{\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}=\mathrm{70} \\ $$$${r}=\mathrm{5}:\:{a}_{\mathrm{5}} =\frac{\mathrm{2}^{\mathrm{5}} ×\mathrm{9}×\mathrm{7}×\mathrm{5}×\mathrm{3}×\mathrm{1}}{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}=\mathrm{252} \\ $$$$…… \\ $$

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