$x^2$ in a different form?
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Is it known that $x^2=$ the sum of all of the numbers underneath it doubled, plus $x$?
Thought of it, thinking about pyramid push ups so $3^2= 2(1)+2(2)+3$,
$5^2= 2(1)+2(2)+2(3)+2(4)+5$
elementary-number-theory
edited Sep 8 at 20:50
Daniel Buck
2,6451625
asked Sep 8 at 7:12
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141
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up vote
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favorite
Is it known that $x^2=$ the sum of all of the numbers underneath it doubled, plus $x$?
Thought of it, thinking about pyramid push ups so $3^2= 2(1)+2(2)+3$,
$5^2= 2(1)+2(2)+2(3)+2(4)+5$
elementary-number-theory
edited Sep 8 at 20:50
Daniel Buck
2,6451625
asked Sep 8 at 7:12
Patch
141
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is it known that $x^2=$ the sum of all of the numbers underneath it doubled, plus $x$?
Thought of it, thinking about pyramid push ups so $3^2= 2(1)+2(2)+3$,
$5^2= 2(1)+2(2)+2(3)+2(4)+5$
elementary-number-theory
edited Sep 8 at 20:50
Daniel Buck
2,6451625
asked Sep 8 at 7:12
Patch
141
Is it known that $x^2=$ the sum of all of the numbers underneath it doubled, plus $x$?
Thought of it, thinking about pyramid push ups so $3^2= 2(1)+2(2)+3$,
$5^2= 2(1)+2(2)+2(3)+2(4)+5$
elementary-number-theory
elementary-number-theory
edited Sep 8 at 20:50
Daniel Buck
2,6451625
asked Sep 8 at 7:12
Patch
141
edited Sep 8 at 20:50
Daniel Buck
2,6451625
asked Sep 8 at 7:12
Patch
141
edited Sep 8 at 20:50
Daniel Buck
2,6451625
edited Sep 8 at 20:50
Daniel Buck
2,6451625
edited Sep 8 at 20:50
Daniel Buck
2,6451625
2,6451625
asked Sep 8 at 7:12
Patch
141
asked Sep 8 at 7:12
Patch
141
asked Sep 8 at 7:12
Patch
141
141
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
Draw dots formed an $ntimes n$ square, and split it to $3$ parts: one diagonal, dots below it, and dots above it.
answered Sep 8 at 7:34
Berci
57.3k23670
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0
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You are asking the following
$$n^2 overset?= 2(1 + 2 + cdots + (n-1)) + n.$$
You have verified it for $n=3$. (You can check $n=2$ as well.)
For a proof by induction, you need to prove the inductive step.
Specifically, if the above holds with $n=k$, can you prove it holds with $n=k+1$?
$$(k+1)^2 = k^2 + 2k + 1 = (2(1+cdots+(k-1)) + k) + 2k + 1 = 2(1+cdots+k) + (k+1) $$
answered Sep 8 at 7:26
angryavian
35.4k12976
What I meant, was has it already been thought of? Please forgive my ignorance.
– Patch
Sep 8 at 7:36
There are lots of this kind of identities, most of them could be found in the e exercise section, for e.g. induction..
– Berci
Sep 8 at 7:39
Thank you. Very much appreciated.
– Patch
Sep 8 at 7:42
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Draw dots formed an $ntimes n$ square, and split it to $3$ parts: one diagonal, dots below it, and dots above it.
answered Sep 8 at 7:34
Berci
57.3k23670
add a comment |Â
up vote
2
down vote
Draw dots formed an $ntimes n$ square, and split it to $3$ parts: one diagonal, dots below it, and dots above it.
answered Sep 8 at 7:34
Berci
57.3k23670
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Draw dots formed an $ntimes n$ square, and split it to $3$ parts: one diagonal, dots below it, and dots above it.
answered Sep 8 at 7:34
Berci
57.3k23670
Draw dots formed an $ntimes n$ square, and split it to $3$ parts: one diagonal, dots below it, and dots above it.
answered Sep 8 at 7:34
Berci
57.3k23670
answered Sep 8 at 7:34
Berci
57.3k23670
answered Sep 8 at 7:34
Berci
57.3k23670
answered Sep 8 at 7:34
Berci
57.3k23670
57.3k23670
add a comment |Â
add a comment |Â
up vote
0
down vote
You are asking the following
$$n^2 overset?= 2(1 + 2 + cdots + (n-1)) + n.$$
You have verified it for $n=3$. (You can check $n=2$ as well.)
For a proof by induction, you need to prove the inductive step.
Specifically, if the above holds with $n=k$, can you prove it holds with $n=k+1$?
$$(k+1)^2 = k^2 + 2k + 1 = (2(1+cdots+(k-1)) + k) + 2k + 1 = 2(1+cdots+k) + (k+1) $$
answered Sep 8 at 7:26
angryavian
35.4k12976
What I meant, was has it already been thought of? Please forgive my ignorance.
– Patch
Sep 8 at 7:36
There are lots of this kind of identities, most of them could be found in the e exercise section, for e.g. induction..
– Berci
Sep 8 at 7:39
Thank you. Very much appreciated.
– Patch
Sep 8 at 7:42
add a comment |Â
up vote
0
down vote
You are asking the following
$$n^2 overset?= 2(1 + 2 + cdots + (n-1)) + n.$$
You have verified it for $n=3$. (You can check $n=2$ as well.)
For a proof by induction, you need to prove the inductive step.
Specifically, if the above holds with $n=k$, can you prove it holds with $n=k+1$?
$$(k+1)^2 = k^2 + 2k + 1 = (2(1+cdots+(k-1)) + k) + 2k + 1 = 2(1+cdots+k) + (k+1) $$
answered Sep 8 at 7:26
angryavian
35.4k12976
What I meant, was has it already been thought of? Please forgive my ignorance.
– Patch
Sep 8 at 7:36
There are lots of this kind of identities, most of them could be found in the e exercise section, for e.g. induction..
– Berci
Sep 8 at 7:39
Thank you. Very much appreciated.
– Patch
Sep 8 at 7:42
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are asking the following
$$n^2 overset?= 2(1 + 2 + cdots + (n-1)) + n.$$
You have verified it for $n=3$. (You can check $n=2$ as well.)
For a proof by induction, you need to prove the inductive step.
Specifically, if the above holds with $n=k$, can you prove it holds with $n=k+1$?
$$(k+1)^2 = k^2 + 2k + 1 = (2(1+cdots+(k-1)) + k) + 2k + 1 = 2(1+cdots+k) + (k+1) $$
answered Sep 8 at 7:26
angryavian
35.4k12976
You are asking the following
$$n^2 overset?= 2(1 + 2 + cdots + (n-1)) + n.$$
You have verified it for $n=3$. (You can check $n=2$ as well.)
For a proof by induction, you need to prove the inductive step.
Specifically, if the above holds with $n=k$, can you prove it holds with $n=k+1$?
$$(k+1)^2 = k^2 + 2k + 1 = (2(1+cdots+(k-1)) + k) + 2k + 1 = 2(1+cdots+k) + (k+1) $$
answered Sep 8 at 7:26
angryavian
35.4k12976
answered Sep 8 at 7:26
angryavian
35.4k12976
answered Sep 8 at 7:26
angryavian
35.4k12976
answered Sep 8 at 7:26
angryavian
35.4k12976
35.4k12976
What I meant, was has it already been thought of? Please forgive my ignorance.
– Patch
Sep 8 at 7:36
There are lots of this kind of identities, most of them could be found in the e exercise section, for e.g. induction..
– Berci
Sep 8 at 7:39
Thank you. Very much appreciated.
– Patch
Sep 8 at 7:42
add a comment |Â
What I meant, was has it already been thought of? Please forgive my ignorance.
– Patch
Sep 8 at 7:36
There are lots of this kind of identities, most of them could be found in the e exercise section, for e.g. induction..
– Berci
Sep 8 at 7:39
Thank you. Very much appreciated.
– Patch
Sep 8 at 7:42
What I meant, was has it already been thought of? Please forgive my ignorance.
– Patch
Sep 8 at 7:36
What I meant, was has it already been thought of? Please forgive my ignorance.
– Patch
Sep 8 at 7:36
There are lots of this kind of identities, most of them could be found in the e exercise section, for e.g. induction..
– Berci
Sep 8 at 7:39
There are lots of this kind of identities, most of them could be found in the e exercise section, for e.g. induction..
– Berci
Sep 8 at 7:39
Thank you. Very much appreciated.
– Patch
Sep 8 at 7:42
Thank you. Very much appreciated.
– Patch
Sep 8 at 7:42
add a comment |Â
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