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
asked Jan 13 at 1:05
user130306user130306
45319
$endgroup$
add a comment |
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
asked Jan 13 at 1:05
user130306user130306
45319
$endgroup$
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
add a comment |
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
asked Jan 13 at 1:05
user130306user130306
45319
$endgroup$
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
asked Jan 13 at 1:05
user130306user130306
45319
asked Jan 13 at 1:05
user130306user130306
45319
edited Jan 17 at 23:36
user130306
asked Jan 13 at 1:05
user130306user130306
45319
asked Jan 13 at 1:05
user130306user130306
45319
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
add a comment |
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
add a comment |
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:
- If $n$ is a multiple of 3, no much more to add.
- If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$
- 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
answered Jan 13 at 1:15
pendermathpendermath
56812
$endgroup$
$begingroup$
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
add a comment |
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 $(3,4)=1$
answered Jan 13 at 2:44
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$begingroup$
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
add a comment |
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:
- If $n$ is a multiple of 3, no much more to add.
- If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$
- 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
answered Jan 13 at 1:15
pendermathpendermath
56812
$endgroup$
$begingroup$
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
add a comment |
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:
- If $n$ is a multiple of 3, no much more to add.
- If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$
- 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
answered Jan 13 at 1:15
pendermathpendermath
56812
$endgroup$
$begingroup$
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
add a comment |
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:
- If $n$ is a multiple of 3, no much more to add.
- If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$
- 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
answered Jan 13 at 1:15
pendermathpendermath
56812
$endgroup$
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:
- If $n$ is a multiple of 3, no much more to add.
- If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$
- 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
answered Jan 13 at 1:15
pendermathpendermath
56812
edited Jan 13 at 1:45
answered Jan 13 at 1:15
pendermathpendermath
56812
answered Jan 13 at 1:15
pendermathpendermath
56812
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
add a comment |
$begingroup$
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
add a comment |
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 $(3,4)=1$
answered Jan 13 at 2:44
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$begingroup$
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
add a comment |
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 $(3,4)=1$
answered Jan 13 at 2:44
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$begingroup$
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
add a comment |
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 $(3,4)=1$
answered Jan 13 at 2:44
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
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 $(3,4)=1$
answered Jan 13 at 2:44
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 13 at 2:44
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 13 at 2:44
lab bhattacharjeelab bhattacharjee
227k15158275
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
add a comment |
$begingroup$
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
add a comment |
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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