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I would like to do a proof by induction. I know there's a few duplicate posts about this question but I dont quite understand the notation they have given while showing an induction proof. If someone could post an induction proof without any complex notation for the product of n consecutive integers, I would really appreciate it.

https://math.stackexchange.com/a/12121/769495 - this answer, how does $(+1)(+2)...(++1) = (k+1) (m+1)...(m+k) + m^{k+1}$ Can anyone expand this notation to show me whats happening?

coolguy
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  • What is the definition of the factorial? – imranfat May 16 '20 at 17:59
  • What is the notation you do not understand? What posts have already answered this question? – amWhy May 16 '20 at 18:01
  • I know that it is simply a way of choosing a number of integers out of another number of integers so it must theoretically always be an integer but a rigorous proof should work too – coolguy May 16 '20 at 18:01
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    The solutions to this duplicate includes proofs by induction. the only notation I see is $m^(k)=m\times (m+1)\times \cdots \times (m+k-1)$ which isn't terribly complicated. If you prefer, you can call it $F(m,k)$ or something like that. – lulu May 16 '20 at 18:02
  • Duplicate – amWhy May 16 '20 at 18:03
  • https://math.stackexchange.com/questions/12065/the-product-of-n-consecutive-integers-is-divisible-by-n-factorial this is the post where i dont understand Nurdin Takenov's answer, especially how (+1)(+2)...(++1) = (k+1) (m+1)...(m+k) + $m^{k+1}$ – coolguy May 16 '20 at 18:03
  • https://math.stackexchange.com/a/12121/769495 here is the answer. how does (+1)(+2)...(++1) = (k+1) (m+1)...(m+k) + $m^{k+1}$ – coolguy May 16 '20 at 18:05
  • The line that you don't understand follows from the distributive law. Just use $(m+k+1)\times A = m\times A + (k+1)\times A$. – lulu May 16 '20 at 18:07
  • what is the A here? is A = k+1? – coolguy May 16 '20 at 18:09
  • No. $A=(m+1)\times \cdots \times (m+k)=(m+1)^{(k)}$. – lulu May 16 '20 at 18:11
  • Maybe it would help to remind you that $m^{(k+1)}$ is NOT $m$ to the $k+1$ power. Rather, it is defined as the product of the integers from $m$ to $m+k$. – Ben W May 16 '20 at 18:17
  • can you also explain how in the same answer, the second term is divided by (+1)! because of induction by – coolguy May 16 '20 at 18:31
  • If you divide thru by $k$ it amounts to one of the common inductive proofs of the integrality of binomial coef's using $\binom{a+b}a = \binom{a+b-1}a + \binom{a+b-1}b$. We already have tens if not hundreds of posts on such. – Bill Dubuque May 16 '20 at 19:28

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I don't consider this a duplicate question because it's asking for more details, or an alternate proof that's easier.

For an easier proof, induction probably is not the way to go. Rather, let's consider the product of all integers from $m+1$ to $m+n$. This can be written as $$ \frac{(m+n)!}{m!} $$ The binomial coefficient $\dbinom{m+n}{n}$ is itself an integer and has the expansion $$ \frac{(m+n)!}{m!(m-[m+n])!} $$ But this is just $$ \dfrac{\dfrac{(m+n)!}{m!}}{n!} $$ QED.

amWhy
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Ben W
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