Question-208332

Question Number 208332 by essaad last updated on 12/Jun/24

Answered by Frix last updated on 12/Jun/24

Π_(k=0) ^n (k+a) =a(a+1)(a+2)...(a+n)=(((a+n)!)/((a−1)!)) ⇒ Π_(k=0) ^n ((k^2 +5k+6)/(k^2 +5k+4)) =Π_(k=0) ^n (((k+2)(k+3))/((k+1)(k+4))) = =(((n+2)!)/((2−1)!))×(((n+3)!)/((3−1)!))×(((1−1)!)/((n+1)!))×(((4−1)!)/((n+4)!))= =(((n+2)!(n+3)!)/((n+1)!(n+4)!))×3= =(((n+1)!(n+2)(n+1)!(n+2)(n+3))/((n+1)!(n+1)!(n+2)(n+3)(n+4)))×3= =((3n+6)/(n+4))

$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\:\left({k}+{a}\right)\:={a}\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right)…\left({a}+{n}\right)=\frac{\left({a}+{n}\right)!}{\left({a}−\mathrm{1}\right)!} \\ $$$$\Rightarrow \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\:\frac{{k}^{\mathrm{2}} +\mathrm{5}{k}+\mathrm{6}}{{k}^{\mathrm{2}} +\mathrm{5}{k}+\mathrm{4}}\:=\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\:\frac{\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{4}\right)}\:= \\ $$$$=\frac{\left({n}+\mathrm{2}\right)!}{\left(\mathrm{2}−\mathrm{1}\right)!}×\frac{\left({n}+\mathrm{3}\right)!}{\left(\mathrm{3}−\mathrm{1}\right)!}×\frac{\left(\mathrm{1}−\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)!}×\frac{\left(\mathrm{4}−\mathrm{1}\right)!}{\left({n}+\mathrm{4}\right)!}= \\ $$$$=\frac{\left({n}+\mathrm{2}\right)!\left({n}+\mathrm{3}\right)!}{\left({n}+\mathrm{1}\right)!\left({n}+\mathrm{4}\right)!}×\mathrm{3}= \\ $$$$=\frac{\left({n}+\mathrm{1}\right)!\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)!\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}{\left({n}+\mathrm{1}\right)!\left({n}+\mathrm{1}\right)!\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)\left({n}+\mathrm{4}\right)}×\mathrm{3}= \\ $$$$=\frac{\mathrm{3}{n}+\mathrm{6}}{{n}+\mathrm{4}} \\ $$

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