Citation/Reference for this Formula [closed]
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Does anyone know any sources that could possibly be used as a citation of some form of the formula below?
$$sum_{n=0}^{k} (a+n)(b-n) = (k+1)ab + frac{k(k+1)}{2}(b-a)+frac{k(k+1)(2k+1)}{6}$$
combinatorics algebraic-combinatorics
asked Jan 13 at 23:23
Jeffrey ShiJeffrey Shi
154
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closed as off-topic by Xander Henderson, Eevee Trainer, max_zorn, Cesareo, metamorphy Jan 14 at 9:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, max_zorn, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Does anyone know any sources that could possibly be used as a citation of some form of the formula below?
$$sum_{n=0}^{k} (a+n)(b-n) = (k+1)ab + frac{k(k+1)}{2}(b-a)+frac{k(k+1)(2k+1)}{6}$$
combinatorics algebraic-combinatorics
asked Jan 13 at 23:23
Jeffrey ShiJeffrey Shi
154
$endgroup$
closed as off-topic by Xander Henderson, Eevee Trainer, max_zorn, Cesareo, metamorphy Jan 14 at 9:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, max_zorn, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I don't know what $b$ is (did you perhaps mean $(a-n)(b-n)$?), but, starting on the left, write $$sum_{n=0}^{k} (a+n)(a-n) = sum_{n=0}^{k} a^2 - sum_{n=0}^{k} n^2, $$ then use standard arguments to get a nice closed form. Pretty much any elementary calculus text should give you the tools you need. Look at the chapter(s) on sequences and series.
$endgroup$
– Xander Henderson
Jan 13 at 23:33
add a comment |
$begingroup$
Does anyone know any sources that could possibly be used as a citation of some form of the formula below?
$$sum_{n=0}^{k} (a+n)(b-n) = (k+1)ab + frac{k(k+1)}{2}(b-a)+frac{k(k+1)(2k+1)}{6}$$
combinatorics algebraic-combinatorics
asked Jan 13 at 23:23
Jeffrey ShiJeffrey Shi
154
$endgroup$
Does anyone know any sources that could possibly be used as a citation of some form of the formula below?
$$sum_{n=0}^{k} (a+n)(b-n) = (k+1)ab + frac{k(k+1)}{2}(b-a)+frac{k(k+1)(2k+1)}{6}$$
combinatorics algebraic-combinatorics
combinatorics algebraic-combinatorics
asked Jan 13 at 23:23
Jeffrey ShiJeffrey Shi
154
asked Jan 13 at 23:23
Jeffrey ShiJeffrey Shi
154
edited Jan 13 at 23:34
Jeffrey Shi
asked Jan 13 at 23:23
Jeffrey ShiJeffrey Shi
154
asked Jan 13 at 23:23
Jeffrey ShiJeffrey Shi
154
asked Jan 13 at 23:23
Jeffrey ShiJeffrey Shi
154
154
closed as off-topic by Xander Henderson, Eevee Trainer, max_zorn, Cesareo, metamorphy Jan 14 at 9:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, max_zorn, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Xander Henderson, Eevee Trainer, max_zorn, Cesareo, metamorphy Jan 14 at 9:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, max_zorn, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I don't know what $b$ is (did you perhaps mean $(a-n)(b-n)$?), but, starting on the left, write $$sum_{n=0}^{k} (a+n)(a-n) = sum_{n=0}^{k} a^2 - sum_{n=0}^{k} n^2, $$ then use standard arguments to get a nice closed form. Pretty much any elementary calculus text should give you the tools you need. Look at the chapter(s) on sequences and series.
$endgroup$
– Xander Henderson
Jan 13 at 23:33
add a comment |
$begingroup$
I don't know what $b$ is (did you perhaps mean $(a-n)(b-n)$?), but, starting on the left, write $$sum_{n=0}^{k} (a+n)(a-n) = sum_{n=0}^{k} a^2 - sum_{n=0}^{k} n^2, $$ then use standard arguments to get a nice closed form. Pretty much any elementary calculus text should give you the tools you need. Look at the chapter(s) on sequences and series.
$endgroup$
– Xander Henderson
Jan 13 at 23:33
$begingroup$
I don't know what $b$ is (did you perhaps mean $(a-n)(b-n)$?), but, starting on the left, write $$sum_{n=0}^{k} (a+n)(a-n) = sum_{n=0}^{k} a^2 - sum_{n=0}^{k} n^2, $$ then use standard arguments to get a nice closed form. Pretty much any elementary calculus text should give you the tools you need. Look at the chapter(s) on sequences and series.
$endgroup$
– Xander Henderson
Jan 13 at 23:33
$begingroup$
I don't know what $b$ is (did you perhaps mean $(a-n)(b-n)$?), but, starting on the left, write $$sum_{n=0}^{k} (a+n)(a-n) = sum_{n=0}^{k} a^2 - sum_{n=0}^{k} n^2, $$ then use standard arguments to get a nice closed form. Pretty much any elementary calculus text should give you the tools you need. Look at the chapter(s) on sequences and series.
$endgroup$
– Xander Henderson
Jan 13 at 23:33
add a comment |
1 Answer
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$begingroup$
Once you've fixed a couple of minor typos that make it false in its current form, it's an essentially-trivial consequence of some very well known expansions, so I'd just cite your favourite sources for those, then prove it. In particular, $(a+n)(b-n) = ab + (b-a)n - n^2$, so
begin{align*}sumlimits_{n=0}^k(a+n)(b-n) &= left(absumlimits_{n=0}^k1right) + (b-a)sumlimits_{n=0}^kn - sumlimits_{n=0}^kn^2
\&=ab(k+1) + (b-a)frac{k(k+1)}{2} - frac{k(k+1)(2k+1)}{6}.end{align*}
(Notice the $b$ replacing your second $a$, and the $-$ instead of $+$ before the final term)
Where the summations are the obvious sum of $k+1$ constants, the standard sum of the first $k$ positive integers, and the standard sum of the first $k$ squares of positive integers.
answered Jan 13 at 23:30
user3482749user3482749
4,3291119
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Once you've fixed a couple of minor typos that make it false in its current form, it's an essentially-trivial consequence of some very well known expansions, so I'd just cite your favourite sources for those, then prove it. In particular, $(a+n)(b-n) = ab + (b-a)n - n^2$, so
begin{align*}sumlimits_{n=0}^k(a+n)(b-n) &= left(absumlimits_{n=0}^k1right) + (b-a)sumlimits_{n=0}^kn - sumlimits_{n=0}^kn^2
\&=ab(k+1) + (b-a)frac{k(k+1)}{2} - frac{k(k+1)(2k+1)}{6}.end{align*}
(Notice the $b$ replacing your second $a$, and the $-$ instead of $+$ before the final term)
Where the summations are the obvious sum of $k+1$ constants, the standard sum of the first $k$ positive integers, and the standard sum of the first $k$ squares of positive integers.
answered Jan 13 at 23:30
user3482749user3482749
4,3291119
$endgroup$
add a comment |
$begingroup$
Once you've fixed a couple of minor typos that make it false in its current form, it's an essentially-trivial consequence of some very well known expansions, so I'd just cite your favourite sources for those, then prove it. In particular, $(a+n)(b-n) = ab + (b-a)n - n^2$, so
begin{align*}sumlimits_{n=0}^k(a+n)(b-n) &= left(absumlimits_{n=0}^k1right) + (b-a)sumlimits_{n=0}^kn - sumlimits_{n=0}^kn^2
\&=ab(k+1) + (b-a)frac{k(k+1)}{2} - frac{k(k+1)(2k+1)}{6}.end{align*}
(Notice the $b$ replacing your second $a$, and the $-$ instead of $+$ before the final term)
Where the summations are the obvious sum of $k+1$ constants, the standard sum of the first $k$ positive integers, and the standard sum of the first $k$ squares of positive integers.
answered Jan 13 at 23:30
user3482749user3482749
4,3291119
$endgroup$
add a comment |
$begingroup$
Once you've fixed a couple of minor typos that make it false in its current form, it's an essentially-trivial consequence of some very well known expansions, so I'd just cite your favourite sources for those, then prove it. In particular, $(a+n)(b-n) = ab + (b-a)n - n^2$, so
begin{align*}sumlimits_{n=0}^k(a+n)(b-n) &= left(absumlimits_{n=0}^k1right) + (b-a)sumlimits_{n=0}^kn - sumlimits_{n=0}^kn^2
\&=ab(k+1) + (b-a)frac{k(k+1)}{2} - frac{k(k+1)(2k+1)}{6}.end{align*}
(Notice the $b$ replacing your second $a$, and the $-$ instead of $+$ before the final term)
Where the summations are the obvious sum of $k+1$ constants, the standard sum of the first $k$ positive integers, and the standard sum of the first $k$ squares of positive integers.
answered Jan 13 at 23:30
user3482749user3482749
4,3291119
$endgroup$
Once you've fixed a couple of minor typos that make it false in its current form, it's an essentially-trivial consequence of some very well known expansions, so I'd just cite your favourite sources for those, then prove it. In particular, $(a+n)(b-n) = ab + (b-a)n - n^2$, so
begin{align*}sumlimits_{n=0}^k(a+n)(b-n) &= left(absumlimits_{n=0}^k1right) + (b-a)sumlimits_{n=0}^kn - sumlimits_{n=0}^kn^2
\&=ab(k+1) + (b-a)frac{k(k+1)}{2} - frac{k(k+1)(2k+1)}{6}.end{align*}
(Notice the $b$ replacing your second $a$, and the $-$ instead of $+$ before the final term)
Where the summations are the obvious sum of $k+1$ constants, the standard sum of the first $k$ positive integers, and the standard sum of the first $k$ squares of positive integers.
answered Jan 13 at 23:30
user3482749user3482749
4,3291119
answered Jan 13 at 23:30
user3482749user3482749
4,3291119
answered Jan 13 at 23:30
user3482749user3482749
4,3291119
answered Jan 13 at 23:30
user3482749user3482749
4,3291119
4,3291119
add a comment |
add a comment |
$begingroup$
I don't know what $b$ is (did you perhaps mean $(a-n)(b-n)$?), but, starting on the left, write $$sum_{n=0}^{k} (a+n)(a-n) = sum_{n=0}^{k} a^2 - sum_{n=0}^{k} n^2, $$ then use standard arguments to get a nice closed form. Pretty much any elementary calculus text should give you the tools you need. Look at the chapter(s) on sequences and series.
$endgroup$
– Xander Henderson
Jan 13 at 23:33