Show that 2|n(n+1) using induction
I tried but im stuck , it still (n+1)(n+2)
Two successive numbers
It's simple using the the methode that n=2k or n=2k+1
Can someone help or give a hint ?
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Two successive numbers; hence, one of them is even. – Martin Peters Sep 07 '15 at 13:51
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6This is a very similar question: Using induction to prove that $2 \mid (n^2 - n)$ for $n\geq 1$. This is a generalization of your question: Proving by strong induction that $\forall n \ge 2, ;\forall d \ge 2$ : $d \mid n(n+1)(n+2)...(n+d-1)$ – Martin Sleziak Sep 07 '15 at 13:56
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Express $(n+1)(n+2)$ as sum of $n(n+1)$ and remainder. Show that remainder is divisible by 2.
Then induction step: if $2|n(n+1)$ then $2|(n+1)(n+2)$.
hvedrung
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It is clear for $n=0$.
Suppose $2 | n(n+1)$. Then either $2 | n$, in which case $2 | n+2$, either $2 | n+1$
Kevin Quirin
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Since $2|n(n+1) \implies n(n+1) \equiv 0 \pmod{2}$. So either $2|n, 2|n+1$.
$(n+1)(n+2) \pmod{2}$ is to be evaluated.
If we took: $2|n \implies n + 1 \equiv 1 \pmod{2} \implies n + 2 \equiv 2 \equiv 0 \pmod{2}$. Hence $(n+1)(n+2)$ is divisible by $2$.
If we took: $2|n+1 \implies n + 1 \equiv 0 \pmod{2}$ then we are done since $(n+1)(n+2) \equiv 0 \pmod{2} \implies 2|(n+1)(n+2)$.
Amad27
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HINT: Suppose $2|k(k+1)$ such that $\frac{k(k+1)}{2}=m$ for some natural numbers k and m where $a_k=k(k+1)$. Then $$a_{k+1}=(k+1)(k+2)$$ and $$a_{k+1}=a_{k}+2(1+k)$$Sub in $a_k$ back into this and rearrange this into a form of $a_{k+1}$ and prove it works for $k=1$. since if its then true for $a_k$ it must then also be true for all k values larger that 1.
Jack Pedley
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