Question Number 113354 by Dwaipayan Shikari last updated on 12/Sep/20
$$\frac{\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}}{\mathrm{2}}−\frac{\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}}{\mathrm{3}}+\frac{\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}}{\mathrm{4}}−\frac{\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}}{\mathrm{5}}+.....{n} \\ $$
Commented by mr W last updated on 13/Sep/20
$$\left(\mathrm{1}−{x}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {x}^{{k}} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {x}^{{k}+\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{{n}} {dx}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \underset{\mathrm{0}} {\int}^{\mathrm{1}} {x}^{{k}+\mathrm{2}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{{n}} {dx}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{{C}_{{k}} ^{{n}} }{{k}+\mathrm{2}} \\ $$$$... \\ $$
Commented by maths mind last updated on 13/Sep/20
$${nice}\:{sir}\:\beta\left({a},{b}\right)=\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{b}−\mathrm{1}} {dx}\:{can}\:{help}\:{you} \\ $$$$ \\ $$$${too}\:{finish}\:{i}\:{hop}\:{you}\:{doing}\:{well}\:{verry}\:{nice}\:{day}\:{sir} \\ $$
Commented by mr W last updated on 14/Sep/20
$${thanks}\:{alot}\:{sir}! \\ $$