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

Answered by Frix last updated on 12/Jun/24

$$\underset{k=0}{\prod ^n}(k+a)=a(a+1)(a+2)...(a+n)=\frac{(a+n)!}{(a-1)!}$$$$\Rightarrow$$$$\underset{k=0}{\prod ^n}\frac{k^2+5k+6}{k^2+5k+4}=\underset{k=0}{\prod ^n}\frac{(k+2)(k+3)}{(k+1)(k+4)}=$$$$=\frac{(n+2)!}{(2-1)!}\times \frac{(n+3)!}{(3-1)!}\times \frac{(1-1)!}{(n+1)!}\times \frac{(4-1)!}{(n+4)!}=$$$$=\frac{(n+2)!(n+3)!}{(n+1)!(n+4)!}\times 3=$$$$=\frac{(n+1)!(n+2)(n+1)!(n+2)(n+3)}{(n+1)!(n+1)!(n+2)(n+3)(n+4)}\times 3=$$$$=\frac{3n+6}{n+4}$$

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