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

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