Prove that if $a \in \mathbb {Z}$ then for every $k \in \mathbb{N}$ it follows that: $k! \mid (a+1) (a+2) \dots (a+k)$

Asked Mar 22 '22 at 20:50

Active Mar 22 '22 at 21:14

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I really don't know how to get around this one, tried to rewrite (a+1)(a+2)...(a+k) as $(a+1)^{k} k!$ and that way k! divides k! and even tho it was close I realized $(a+1)^{k} k!$ was not really equal to (a+1)(a+2)...(a+k). So now I don't have the slightest clue on how to procede

edited Mar 22 '22 at 21:14

asked Mar 22 '22 at 20:50

Eduardo Abraham Vivaldo Daza

5

Does this answer your question? The product of $n$ consecutive integers is divisible by $n$ factorial. Found by using Approach0. – John Omielan Mar 22 '22 at 20:52
Write the right-hand side as the binomial coefficient and you should be able to see it – KaiL1ng Mar 22 '22 at 21:03

Prove that if $a \in \mathbb {Z}$ then for every $k \in \mathbb{N}$ it follows that: $k! \mid (a+1) (a+2) \dots (a+k)$

0 Answers0