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

408111




408111








  • 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












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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

57612




57612












  • $begingroup$
    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

229k15159279




229k15159279












  • $begingroup$
    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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.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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