How can I prove












0












$begingroup$


I'm in high school so I don't have a lot of proof techniques.I also don't know modular arithmetic so I can't use that either. Any help is appreciated, thank you!






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    as far as 2, you can calculate the thing for cases $n=2k$ and $n=2k+1.$ In both cases the product is even. Similar for 3, cases are $n = 3w,$ $n=3w+1,$ finally $n=3w+2.$ Just calculation, but you do need to do that much.
    $endgroup$
    – Will Jagy
    Jan 13 at 1:09








  • 3




    $begingroup$
    If you know the sum of squares formula, you'll know that $(2n + 1)(n + 1)n$ is $6$ times the sum of the first $n$ squares.
    $endgroup$
    – Theo Bendit
    Jan 13 at 1:14
















0












$begingroup$


I'm in high school so I don't have a lot of proof techniques.I also don't know modular arithmetic so I can't use that either. Any help is appreciated, thank you!






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    as far as 2, you can calculate the thing for cases $n=2k$ and $n=2k+1.$ In both cases the product is even. Similar for 3, cases are $n = 3w,$ $n=3w+1,$ finally $n=3w+2.$ Just calculation, but you do need to do that much.
    $endgroup$
    – Will Jagy
    Jan 13 at 1:09








  • 3




    $begingroup$
    If you know the sum of squares formula, you'll know that $(2n + 1)(n + 1)n$ is $6$ times the sum of the first $n$ squares.
    $endgroup$
    – Theo Bendit
    Jan 13 at 1:14














0












0








0





$begingroup$


I'm in high school so I don't have a lot of proof techniques.I also don't know modular arithmetic so I can't use that either. Any help is appreciated, thank you!






algebra-precalculus







share|cite|improve this question











$endgroup$




I'm in high school so I don't have a lot of proof techniques.I also don't know modular arithmetic so I can't use that either. Any help is appreciated, thank you!






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 23:36







user130306

















asked Jan 13 at 1:05









user130306user130306

45319




45319








  • 1




    $begingroup$
    as far as 2, you can calculate the thing for cases $n=2k$ and $n=2k+1.$ In both cases the product is even. Similar for 3, cases are $n = 3w,$ $n=3w+1,$ finally $n=3w+2.$ Just calculation, but you do need to do that much.
    $endgroup$
    – Will Jagy
    Jan 13 at 1:09








  • 3




    $begingroup$
    If you know the sum of squares formula, you'll know that $(2n + 1)(n + 1)n$ is $6$ times the sum of the first $n$ squares.
    $endgroup$
    – Theo Bendit
    Jan 13 at 1:14














  • 1




    $begingroup$
    as far as 2, you can calculate the thing for cases $n=2k$ and $n=2k+1.$ In both cases the product is even. Similar for 3, cases are $n = 3w,$ $n=3w+1,$ finally $n=3w+2.$ Just calculation, but you do need to do that much.
    $endgroup$
    – Will Jagy
    Jan 13 at 1:09








  • 3




    $begingroup$
    If you know the sum of squares formula, you'll know that $(2n + 1)(n + 1)n$ is $6$ times the sum of the first $n$ squares.
    $endgroup$
    – Theo Bendit
    Jan 13 at 1:14








1




1




$begingroup$
as far as 2, you can calculate the thing for cases $n=2k$ and $n=2k+1.$ In both cases the product is even. Similar for 3, cases are $n = 3w,$ $n=3w+1,$ finally $n=3w+2.$ Just calculation, but you do need to do that much.
$endgroup$
– Will Jagy
Jan 13 at 1:09






$begingroup$
as far as 2, you can calculate the thing for cases $n=2k$ and $n=2k+1.$ In both cases the product is even. Similar for 3, cases are $n = 3w,$ $n=3w+1,$ finally $n=3w+2.$ Just calculation, but you do need to do that much.
$endgroup$
– Will Jagy
Jan 13 at 1:09






3




3




$begingroup$
If you know the sum of squares formula, you'll know that $(2n + 1)(n + 1)n$ is $6$ times the sum of the first $n$ squares.
$endgroup$
– Theo Bendit
Jan 13 at 1:14




$begingroup$
If you know the sum of squares formula, you'll know that $(2n + 1)(n + 1)n$ is $6$ times the sum of the first $n$ squares.
$endgroup$
– Theo Bendit
Jan 13 at 1:14










2 Answers
2






active

oldest

votes


















5












$begingroup$

First note that either $n$ or $n+1$ is a multiple of 2. Then, there are three cases to see that the number is a multiple of 3 as well:




  1. If $n$ is a multiple of 3, no much more to add.

  2. If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$

  3. If $n$ is a multiple of 3 minus 1, then $n+1$ is a multiple of 3.


In any case, $(2n+1)(n+1)n$ is multiple of 6






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Right idea but not quite right. The first sentence is wrong when $n=6$. Settle divisibility by $2$ by looking at $n(n+1)$.
    $endgroup$
    – Ethan Bolker
    Jan 13 at 1:18










  • $begingroup$
    Corrected. Thx!
    $endgroup$
    – pendermath
    Jan 13 at 1:23










  • $begingroup$
    You don't need anything about $n$ mod $3$ to know $n(n+1)$ is even. Do that first. Then just worry about $3$.
    $endgroup$
    – Ethan Bolker
    Jan 13 at 1:42










  • $begingroup$
    Good point. My solution is effective, it could be more efficient. I changed it following your suggestions
    $endgroup$
    – pendermath
    Jan 13 at 1:43





















2












$begingroup$

As $n(n+1)$ is even, it is sufficient to show $3$ divides $(2n+1)n(n+1)$



Now $2n(2n+1)(2n+2)$ is divisible by $3,$ being product of three consecutive integers for any integer $n$



$3$ will divide $dfrac{2n(2n+1)(2n+2)}{2cdot2}$ as $(3,4)=1$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I really like that trick, using a product of three consecutive numbers. I'll remember that. +1
    $endgroup$
    – Theo Bendit
    Jan 13 at 11:48










  • $begingroup$
    @Theo, math.stackexchange.com/questions/12065/…
    $endgroup$
    – lab bhattacharjee
    Jan 13 at 12:04










  • $begingroup$
    I was more referring to the trick of relating $(2n + 1)n(n + 1)$ to $(2n)(2n + 1)(2n + 2)$. I never thought of explaining the divisibility by $3$ in this way.
    $endgroup$
    – Theo Bendit
    Jan 13 at 12:34











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

First note that either $n$ or $n+1$ is a multiple of 2. Then, there are three cases to see that the number is a multiple of 3 as well:




  1. If $n$ is a multiple of 3, no much more to add.

  2. If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$

  3. If $n$ is a multiple of 3 minus 1, then $n+1$ is a multiple of 3.


In any case, $(2n+1)(n+1)n$ is multiple of 6






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Right idea but not quite right. The first sentence is wrong when $n=6$. Settle divisibility by $2$ by looking at $n(n+1)$.
    $endgroup$
    – Ethan Bolker
    Jan 13 at 1:18










  • $begingroup$
    Corrected. Thx!
    $endgroup$
    – pendermath
    Jan 13 at 1:23










  • $begingroup$
    You don't need anything about $n$ mod $3$ to know $n(n+1)$ is even. Do that first. Then just worry about $3$.
    $endgroup$
    – Ethan Bolker
    Jan 13 at 1:42










  • $begingroup$
    Good point. My solution is effective, it could be more efficient. I changed it following your suggestions
    $endgroup$
    – pendermath
    Jan 13 at 1:43


















5












$begingroup$

First note that either $n$ or $n+1$ is a multiple of 2. Then, there are three cases to see that the number is a multiple of 3 as well:




  1. If $n$ is a multiple of 3, no much more to add.

  2. If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$

  3. If $n$ is a multiple of 3 minus 1, then $n+1$ is a multiple of 3.


In any case, $(2n+1)(n+1)n$ is multiple of 6






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Right idea but not quite right. The first sentence is wrong when $n=6$. Settle divisibility by $2$ by looking at $n(n+1)$.
    $endgroup$
    – Ethan Bolker
    Jan 13 at 1:18










  • $begingroup$
    Corrected. Thx!
    $endgroup$
    – pendermath
    Jan 13 at 1:23










  • $begingroup$
    You don't need anything about $n$ mod $3$ to know $n(n+1)$ is even. Do that first. Then just worry about $3$.
    $endgroup$
    – Ethan Bolker
    Jan 13 at 1:42










  • $begingroup$
    Good point. My solution is effective, it could be more efficient. I changed it following your suggestions
    $endgroup$
    – pendermath
    Jan 13 at 1:43
















5












5








5





$begingroup$

First note that either $n$ or $n+1$ is a multiple of 2. Then, there are three cases to see that the number is a multiple of 3 as well:




  1. If $n$ is a multiple of 3, no much more to add.

  2. If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$

  3. If $n$ is a multiple of 3 minus 1, then $n+1$ is a multiple of 3.


In any case, $(2n+1)(n+1)n$ is multiple of 6






share|cite|improve this answer











$endgroup$



First note that either $n$ or $n+1$ is a multiple of 2. Then, there are three cases to see that the number is a multiple of 3 as well:




  1. If $n$ is a multiple of 3, no much more to add.

  2. If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$

  3. If $n$ is a multiple of 3 minus 1, then $n+1$ is a multiple of 3.


In any case, $(2n+1)(n+1)n$ is multiple of 6







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 at 1:45

























answered Jan 13 at 1:15









pendermathpendermath

56812




56812












  • $begingroup$
    Right idea but not quite right. The first sentence is wrong when $n=6$. Settle divisibility by $2$ by looking at $n(n+1)$.
    $endgroup$
    – Ethan Bolker
    Jan 13 at 1:18










  • $begingroup$
    Corrected. Thx!
    $endgroup$
    – pendermath
    Jan 13 at 1:23










  • $begingroup$
    You don't need anything about $n$ mod $3$ to know $n(n+1)$ is even. Do that first. Then just worry about $3$.
    $endgroup$
    – Ethan Bolker
    Jan 13 at 1:42










  • $begingroup$
    Good point. My solution is effective, it could be more efficient. I changed it following your suggestions
    $endgroup$
    – pendermath
    Jan 13 at 1:43




















  • $begingroup$
    Right idea but not quite right. The first sentence is wrong when $n=6$. Settle divisibility by $2$ by looking at $n(n+1)$.
    $endgroup$
    – Ethan Bolker
    Jan 13 at 1:18










  • $begingroup$
    Corrected. Thx!
    $endgroup$
    – pendermath
    Jan 13 at 1:23










  • $begingroup$
    You don't need anything about $n$ mod $3$ to know $n(n+1)$ is even. Do that first. Then just worry about $3$.
    $endgroup$
    – Ethan Bolker
    Jan 13 at 1:42










  • $begingroup$
    Good point. My solution is effective, it could be more efficient. I changed it following your suggestions
    $endgroup$
    – pendermath
    Jan 13 at 1:43


















$begingroup$
Right idea but not quite right. The first sentence is wrong when $n=6$. Settle divisibility by $2$ by looking at $n(n+1)$.
$endgroup$
– Ethan Bolker
Jan 13 at 1:18




$begingroup$
Right idea but not quite right. The first sentence is wrong when $n=6$. Settle divisibility by $2$ by looking at $n(n+1)$.
$endgroup$
– Ethan Bolker
Jan 13 at 1:18












$begingroup$
Corrected. Thx!
$endgroup$
– pendermath
Jan 13 at 1:23




$begingroup$
Corrected. Thx!
$endgroup$
– pendermath
Jan 13 at 1:23












$begingroup$
You don't need anything about $n$ mod $3$ to know $n(n+1)$ is even. Do that first. Then just worry about $3$.
$endgroup$
– Ethan Bolker
Jan 13 at 1:42




$begingroup$
You don't need anything about $n$ mod $3$ to know $n(n+1)$ is even. Do that first. Then just worry about $3$.
$endgroup$
– Ethan Bolker
Jan 13 at 1:42












$begingroup$
Good point. My solution is effective, it could be more efficient. I changed it following your suggestions
$endgroup$
– pendermath
Jan 13 at 1:43






$begingroup$
Good point. My solution is effective, it could be more efficient. I changed it following your suggestions
$endgroup$
– pendermath
Jan 13 at 1:43













2












$begingroup$

As $n(n+1)$ is even, it is sufficient to show $3$ divides $(2n+1)n(n+1)$



Now $2n(2n+1)(2n+2)$ is divisible by $3,$ being product of three consecutive integers for any integer $n$



$3$ will divide $dfrac{2n(2n+1)(2n+2)}{2cdot2}$ as $(3,4)=1$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I really like that trick, using a product of three consecutive numbers. I'll remember that. +1
    $endgroup$
    – Theo Bendit
    Jan 13 at 11:48










  • $begingroup$
    @Theo, math.stackexchange.com/questions/12065/…
    $endgroup$
    – lab bhattacharjee
    Jan 13 at 12:04










  • $begingroup$
    I was more referring to the trick of relating $(2n + 1)n(n + 1)$ to $(2n)(2n + 1)(2n + 2)$. I never thought of explaining the divisibility by $3$ in this way.
    $endgroup$
    – Theo Bendit
    Jan 13 at 12:34
















2












$begingroup$

As $n(n+1)$ is even, it is sufficient to show $3$ divides $(2n+1)n(n+1)$



Now $2n(2n+1)(2n+2)$ is divisible by $3,$ being product of three consecutive integers for any integer $n$



$3$ will divide $dfrac{2n(2n+1)(2n+2)}{2cdot2}$ as $(3,4)=1$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I really like that trick, using a product of three consecutive numbers. I'll remember that. +1
    $endgroup$
    – Theo Bendit
    Jan 13 at 11:48










  • $begingroup$
    @Theo, math.stackexchange.com/questions/12065/…
    $endgroup$
    – lab bhattacharjee
    Jan 13 at 12:04










  • $begingroup$
    I was more referring to the trick of relating $(2n + 1)n(n + 1)$ to $(2n)(2n + 1)(2n + 2)$. I never thought of explaining the divisibility by $3$ in this way.
    $endgroup$
    – Theo Bendit
    Jan 13 at 12:34














2












2








2





$begingroup$

As $n(n+1)$ is even, it is sufficient to show $3$ divides $(2n+1)n(n+1)$



Now $2n(2n+1)(2n+2)$ is divisible by $3,$ being product of three consecutive integers for any integer $n$



$3$ will divide $dfrac{2n(2n+1)(2n+2)}{2cdot2}$ as $(3,4)=1$






share|cite|improve this answer









$endgroup$



As $n(n+1)$ is even, it is sufficient to show $3$ divides $(2n+1)n(n+1)$



Now $2n(2n+1)(2n+2)$ is divisible by $3,$ being product of three consecutive integers for any integer $n$



$3$ will divide $dfrac{2n(2n+1)(2n+2)}{2cdot2}$ as $(3,4)=1$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 at 2:44









lab bhattacharjeelab bhattacharjee

227k15158275




227k15158275












  • $begingroup$
    I really like that trick, using a product of three consecutive numbers. I'll remember that. +1
    $endgroup$
    – Theo Bendit
    Jan 13 at 11:48










  • $begingroup$
    @Theo, math.stackexchange.com/questions/12065/…
    $endgroup$
    – lab bhattacharjee
    Jan 13 at 12:04










  • $begingroup$
    I was more referring to the trick of relating $(2n + 1)n(n + 1)$ to $(2n)(2n + 1)(2n + 2)$. I never thought of explaining the divisibility by $3$ in this way.
    $endgroup$
    – Theo Bendit
    Jan 13 at 12:34


















  • $begingroup$
    I really like that trick, using a product of three consecutive numbers. I'll remember that. +1
    $endgroup$
    – Theo Bendit
    Jan 13 at 11:48










  • $begingroup$
    @Theo, math.stackexchange.com/questions/12065/…
    $endgroup$
    – lab bhattacharjee
    Jan 13 at 12:04










  • $begingroup$
    I was more referring to the trick of relating $(2n + 1)n(n + 1)$ to $(2n)(2n + 1)(2n + 2)$. I never thought of explaining the divisibility by $3$ in this way.
    $endgroup$
    – Theo Bendit
    Jan 13 at 12:34
















$begingroup$
I really like that trick, using a product of three consecutive numbers. I'll remember that. +1
$endgroup$
– Theo Bendit
Jan 13 at 11:48




$begingroup$
I really like that trick, using a product of three consecutive numbers. I'll remember that. +1
$endgroup$
– Theo Bendit
Jan 13 at 11:48












$begingroup$
@Theo, math.stackexchange.com/questions/12065/…
$endgroup$
– lab bhattacharjee
Jan 13 at 12:04




$begingroup$
@Theo, math.stackexchange.com/questions/12065/…
$endgroup$
– lab bhattacharjee
Jan 13 at 12:04












$begingroup$
I was more referring to the trick of relating $(2n + 1)n(n + 1)$ to $(2n)(2n + 1)(2n + 2)$. I never thought of explaining the divisibility by $3$ in this way.
$endgroup$
– Theo Bendit
Jan 13 at 12:34




$begingroup$
I was more referring to the trick of relating $(2n + 1)n(n + 1)$ to $(2n)(2n + 1)(2n + 2)$. I never thought of explaining the divisibility by $3$ in this way.
$endgroup$
– Theo Bendit
Jan 13 at 12:34


















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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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