Question Number 125257 by Schwartzman94 last updated on 09/Dec/20
$${S}=\frac{{C}_{{n}} ^{\mathrm{0}} }{{C}_{{n}+\mathrm{2}} ^{\mathrm{1}} }+\frac{{C}_{{n}} ^{\mathrm{1}} }{{C}_{{n}+\mathrm{2}} ^{\mathrm{2}} }+…+\frac{{C}_{{n}} ^{{n}} }{{C}_{{n}+\mathrm{2}} ^{{n}+\mathrm{1}} } \\ $$
Commented by mr W last updated on 09/Dec/20
$${S}=\frac{{n}+\mathrm{3}}{\mathrm{6}} \\ $$
Answered by Olaf last updated on 09/Dec/20
![S = Σ_(k=0) ^(k=n) (C_n ^k /C_(n+2) ^(k+1) ) S = Σ_(k=0) ^(k=n) (((n!)/(k!(n−k)!))/(((n+2)!)/((k+1)!(n−k+1)!))) S = Σ_(k=0) ^(k=n) (((n!)/(k!(n−k)!))/((n!(n+1)(n+2))/(k!(k+1)(n−k)!(n−k+1)))) S = Σ_(k=0) ^(k=n) (1/(((n+1)(n+2))/((k+1)(n−k+1)))) S = (1/((n+1)(n+2)))Σ_(k=0) ^(k=n) (k+1)(n−k+1) S = (1/((n+1)(n+2)))Σ_(k=1) ^(k=n+1) k(n+2−k) S = (1/((n+1)(n+2)))[(n+2)Σ_(k=1) ^(k=n+1) k−Σ_(k=1) ^(k=n+1) k^2 ] S = (1/((n+1)(n+2)))[(n+2)(((n+1)(n+2))/2)−(((n+1)(n+2)(2n+3))/6)] S = ((3(n+2)−(2n+3))/6) S = ((n+3)/6)](https://www.tinkutara.com/question/Q125285.png)
$$ \\ $$$$\mathrm{S}\:=\:\underset{{k}=\mathrm{0}} {\overset{{k}={n}} {\sum}}\frac{\mathrm{C}_{{n}} ^{{k}} }{\mathrm{C}_{{n}+\mathrm{2}} ^{{k}+\mathrm{1}} } \\ $$$$\mathrm{S}\:=\:\underset{{k}=\mathrm{0}} {\overset{{k}={n}} {\sum}}\frac{\frac{{n}!}{{k}!\left({n}−{k}\right)!}}{\frac{\left({n}+\mathrm{2}\right)!}{\left({k}+\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!}} \\ $$$$\mathrm{S}\:=\:\underset{{k}=\mathrm{0}} {\overset{{k}={n}} {\sum}}\frac{\frac{{n}!}{{k}!\left({n}−{k}\right)!}}{\frac{{n}!\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{{k}!\left({k}+\mathrm{1}\right)\left({n}−{k}\right)!\left({n}−{k}+\mathrm{1}\right)}} \\ $$$$\mathrm{S}\:=\:\underset{{k}=\mathrm{0}} {\overset{{k}={n}} {\sum}}\frac{\mathrm{1}}{\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\left({k}+\mathrm{1}\right)\left({n}−{k}+\mathrm{1}\right)}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\underset{{k}=\mathrm{0}} {\overset{{k}={n}} {\sum}}\left({k}+\mathrm{1}\right)\left({n}−{k}+\mathrm{1}\right) \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\underset{{k}=\mathrm{1}} {\overset{{k}={n}+\mathrm{1}} {\sum}}{k}\left({n}+\mathrm{2}−{k}\right) \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\left[\left({n}+\mathrm{2}\right)\underset{{k}=\mathrm{1}} {\overset{{k}={n}+\mathrm{1}} {\sum}}{k}−\underset{{k}=\mathrm{1}} {\overset{{k}={n}+\mathrm{1}} {\sum}}{k}^{\mathrm{2}} \right] \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\left[\left({n}+\mathrm{2}\right)\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{2}}−\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}{\mathrm{6}}\right] \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{3}\left({n}+\mathrm{2}\right)−\left(\mathrm{2}{n}+\mathrm{3}\right)}{\mathrm{6}} \\ $$$$\mathrm{S}\:=\:\frac{{n}+\mathrm{3}}{\mathrm{6}} \\ $$