Question Number 36437 by prof Abdo imad last updated on 02/Jun/18
$${simplify}\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\:\frac{{C}_{{n}} ^{{k}} }{{k}+\mathrm{1}} \\ $$
Commented by prakash jain last updated on 02/Jun/18
$$\frac{\mathrm{1}}{{k}+\mathrm{1}}\:^{{n}} {C}_{{k}} =\left(\frac{\mathrm{1}}{{k}+\mathrm{1}}\right)\frac{{n}!}{\left({n}−{k}\right)!{k}!} \\ $$$$=\left(\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\right)\frac{\left({n}+\mathrm{1}\right)!}{\left({n}−{k}\right)!\left({k}+\mathrm{1}\right)!} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{k}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}+\mathrm{1}} {C}_{{k}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$
Commented by prof Abdo imad last updated on 03/Jun/18
$${let}\:{consider}\:{the}\:{polynom}\:\:{P}\left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\frac{{x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}} \\ $$$${we}\:{have}\:\:\frac{{dP}}{{dx}}\left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} =\:\left({x}+\mathrm{1}\right)^{{n}} \:\Rightarrow \\ $$$${P}\left({x}\right)\:=\:\int\:\left({x}+\mathrm{1}\right)^{{n}} \:+{c}\:=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} \:+{c}\: \\ $$$${P}\left(\mathrm{0}\right)=\mathrm{0}\:=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+{c}\:\Rightarrow{c}\:=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow \\ $$$${P}\left({x}\right)\:=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\left\{\:\left({x}+\mathrm{1}\right)^{{n}} \:−\mathrm{1}\right\}\:\:{and} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{{C}_{{n}} ^{{k}} }{{k}+\mathrm{1}}\:=\:{P}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{2}^{{n}} \:−\mathrm{1}}{{n}+\mathrm{1}\:}\:\:. \\ $$
Commented by prof Abdo imad last updated on 03/Jun/18
$${error}\:{at}\:{the}\:{final}\:{lines} \\ $$$${P}\left({x}\right)=\frac{\mathrm{1}}{{n}+\mathrm{1}}\left\{\:\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} \:−\mathrm{1}\right\}\:{and} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{{C}_{{n}} ^{{k}} }{{k}+\mathrm{1}}\:={P}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{2}^{{n}+\mathrm{1}} \:−\mathrm{1}}{{n}+\mathrm{1}}\:\bigstar \\ $$$$ \\ $$