Question Number 25969 by soyebshaikh41@gmail.com last updated on 17/Dec/17
$$\mathrm{C}_{\mathrm{0}} +\mathrm{2C}_{\mathrm{1}} +\mathrm{3C}_{\mathrm{2}} +..........+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{C}_{\mathrm{n}} =\left(\mathrm{n}+\mathrm{2}\right)\mathrm{2}^{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{using}\:\:\mathrm{Bionomial}\:\mathrm{teorm} \\ $$
Commented by moxhix last updated on 17/Dec/17
$${show}\:_{{n}} {C}_{\mathrm{0}} +\mathrm{2}\:_{{n}} {C}_{\mathrm{1}} +\mathrm{3}\:_{{n}} {C}_{\mathrm{2}} +...+\left({n}+\mathrm{1}\right)\:_{{n}} {C}_{{n}} =\left({n}+\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{1}} \\ $$$${put}\:{f}\left({x}\right)={x}\left(\mathrm{1}+{x}\right)^{{n}} \\ $$$$ \\ $$$${f}\:'\left({x}\right)=\left(\mathrm{1}+{x}\right)^{{n}} +{nx}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} =\left({n}+\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} \\ $$$${f}\:'\left(\mathrm{1}\right)=\left({n}+\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{1}} \\ $$$$ \\ $$$${f}\left({x}\right)={x}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:_{{n}} {C}_{{k}} {x}^{{k}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:_{{n}} {C}_{{k}} {x}^{{k}+\mathrm{1}} \\ $$$${f}\:'\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:_{{n}} {C}_{{k}} \left({k}+\mathrm{1}\right){x}^{{k}} \\ $$$${f}\:'\left(\mathrm{1}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:_{{n}} {C}_{{k}} \left({k}+\mathrm{1}\right) \\ $$$$ \\ $$$$\therefore\:_{{n}} {C}_{\mathrm{0}} +\mathrm{2}\:_{{n}} {C}_{\mathrm{1}} +\mathrm{3}\:_{{n}} {C}_{\mathrm{2}} +...+\left({n}+\mathrm{1}\right)\:_{{n}} {C}_{{n}} =\left({n}+\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{1}} \\ $$