Question Number 22177 by Tinkutara last updated on 12/Oct/17
$$\frac{{C}_{\mathrm{0}} }{\mathrm{2}}\:−\:\frac{{C}_{\mathrm{1}} }{\mathrm{3}}\:+\:\frac{{C}_{\mathrm{2}} }{\mathrm{4}}\:−\:\frac{{C}_{\mathrm{3}} }{\mathrm{5}}\:+\:.......... \\ $$
Answered by ajfour last updated on 12/Oct/17
$${x}\left(\mathrm{1}−{x}\right)^{{n}} ={C}_{\mathrm{0}} {x}−{C}_{\mathrm{1}} {x}^{\mathrm{2}} +{C}_{\mathrm{2}} {x}^{\mathrm{3}} +.. \\ $$$$\int_{\mathrm{0}} ^{\:\:\mathrm{1}} {x}\left(\mathrm{1}−{x}\right)^{{n}} {dx}=\frac{{C}_{\mathrm{0}} }{\mathrm{2}}−\frac{{C}_{\mathrm{1}} }{\mathrm{3}}+\frac{{C}_{\mathrm{2}} }{\mathrm{4}}−... \\ $$$${for}\:{the}\:{integral}\:{let}\:\:\mathrm{1}−{x}={t} \\ $$$$\frac{{C}_{\mathrm{0}} }{\mathrm{2}}−\frac{{C}_{\mathrm{1}} }{\mathrm{3}}+\frac{{C}_{\mathrm{2}} }{\mathrm{4}}−...=\:−\int_{\mathrm{1}} ^{\:\:\mathrm{0}} \left(\mathrm{1}−{t}\right){t}^{{n}} {dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \left({t}^{{n}} −{t}^{{n}+\mathrm{1}} \right){dt} \\ $$$$\:\:\:\:\:=\left(\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}−\frac{{t}^{{n}+\mathrm{2}} }{{n}+\mathrm{2}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\:=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:. \\ $$
Commented by Tinkutara last updated on 13/Oct/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$