Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

prove by induction that $n(n+1)(n+2)(n+3)$ is an integer multiple of $24$

Let $P(n)$ be the proposition we want to prov, ie: $P(n):=24 \mid(n)(n+1)(n+2)(n+3)$

For $P(1)$ we have: $24 \mid(1)(1+1)(1+2)(1+3)\implies6 \mid(1)(2)(3)(4)\implies24 \mid24$, so $P(1)$is true.

For $P(2)$ we have: $24 \mid(2)(2+1)(2+2)(2+3)\implies6 \mid(2)(3)(4)(5)\implies24 \mid120$, so $P(1)$is true

Inductive Hypothesis: Let $n=k$ and we assume that $P(k):=24\mid k(k+1)(k+2)(k+3)$ is true.

Inductive Step: $$(k+1)(k+2)(k+3)(k+4)$$ $$k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)$$ Using the assumption of $P(k) \implies \exists a\in \mathbb Z$, such that, $(k+1)(k+2)(k+3)=24\cdot a$

so: $$=24\cdot a +4(k+1)(k+2)(k+3)$$

share|improve this question
    
For the obvious induction to work, we need the fact that the product of $3$ consecutive integers is divisible by $6$. If we are going to do that by induction, we need the fact that the product of $2$ consecutive integers is divisible by $2$. But we can also give non-induction arguments for these facts, (and also for the fact about $4$ consecutives). One can also do a double induction and prove that the product of $m$ consecutives is divisible by $m!$. –  André Nicolas Jan 9 at 7:31

1 Answer 1

up vote 3 down vote accepted

Continuing from where you left, we just need to prove that for $n=k+1$, $P(n)$ is an integer multiple of $24$.

$$P(k+1) = (k+1)(k+2)(k+3)(k+4)$$ $$P(k+1) = k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)$$ $1st$ term on right hand side is $P(k)$ which is an integer multiple of $24$ from your inductive hypothesis.

$2nd$ term on the right hand side has a product of $3$ consecutive integers and hence divisible by $6$. So on a whole divisible by $4*6=24$.

On a whole the right hand side is divisible by $24$. Hence $P(k+1)$ is an integral multiple of $24$.

share|improve this answer
    
thank you for the last comment, i didn´t had in mind that $6\mid n(n+1)(n+2)$ –  Victor Francisco Salazar Garci Jan 9 at 7:27
    
@VictorFranciscoSalazarGarci No Problem. If you are satisfied with any of the answers you get for your question accept the answer so that others do not waste their time by posting again. –  lsp Jan 9 at 7:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.