What would be the closed form of $ncdot n + (n-2)cdot(n-2) + (n-4)cdot(n-4) + dots$?
$begingroup$
As part of solving this problem , I came up with the following expression:
$$text{ans} = ncdot n + (n-2)cdot (n-2) + (n-4)cdot (n-4) + dots$$
So I am just running a loop to calculate the ans. However, since there is a pattern in the form of a series, I am wondering if a closed form exists for it? Please also show the steps to obtain so that I can learn, in general, how to approach such problems for finding closed forms.
EDIT: series goes till 2 if n is even, and till 1 if n is odd.
sequences-and-series closed-form
asked Aug 17 '18 at 9:54
user3243499user3243499
192114
$endgroup$
|
show 3 more comments
$begingroup$
As part of solving this problem , I came up with the following expression:
$$text{ans} = ncdot n + (n-2)cdot (n-2) + (n-4)cdot (n-4) + dots$$
So I am just running a loop to calculate the ans. However, since there is a pattern in the form of a series, I am wondering if a closed form exists for it? Please also show the steps to obtain so that I can learn, in general, how to approach such problems for finding closed forms.
EDIT: series goes till 2 if n is even, and till 1 if n is odd.
sequences-and-series closed-form
asked Aug 17 '18 at 9:54
user3243499user3243499
192114
$endgroup$
$begingroup$
Do you know the formula for the sum of the first $m$ squares?
$endgroup$
– Arnaud D.
Aug 17 '18 at 9:55
$begingroup$
yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
$endgroup$
– user3243499
Aug 17 '18 at 10:00
$begingroup$
It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
$endgroup$
– Claude Leibovici
Aug 17 '18 at 10:03
$begingroup$
Where exactly does it stop?
$endgroup$
– Arnaud Mortier
Aug 17 '18 at 10:03
$begingroup$
@user3243499: no, it's the sum of the squares of the same parity.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:04
|
show 3 more comments
$begingroup$
As part of solving this problem , I came up with the following expression:
$$text{ans} = ncdot n + (n-2)cdot (n-2) + (n-4)cdot (n-4) + dots$$
So I am just running a loop to calculate the ans. However, since there is a pattern in the form of a series, I am wondering if a closed form exists for it? Please also show the steps to obtain so that I can learn, in general, how to approach such problems for finding closed forms.
EDIT: series goes till 2 if n is even, and till 1 if n is odd.
sequences-and-series closed-form
asked Aug 17 '18 at 9:54
user3243499user3243499
192114
$endgroup$
As part of solving this problem , I came up with the following expression:
$$text{ans} = ncdot n + (n-2)cdot (n-2) + (n-4)cdot (n-4) + dots$$
So I am just running a loop to calculate the ans. However, since there is a pattern in the form of a series, I am wondering if a closed form exists for it? Please also show the steps to obtain so that I can learn, in general, how to approach such problems for finding closed forms.
EDIT: series goes till 2 if n is even, and till 1 if n is odd.
sequences-and-series closed-form
sequences-and-series closed-form
asked Aug 17 '18 at 9:54
user3243499user3243499
192114
asked Aug 17 '18 at 9:54
user3243499user3243499
192114
edited Aug 17 '18 at 10:12
user3243499
asked Aug 17 '18 at 9:54
user3243499user3243499
192114
asked Aug 17 '18 at 9:54
user3243499user3243499
192114
asked Aug 17 '18 at 9:54
user3243499user3243499
192114
192114
$begingroup$
Do you know the formula for the sum of the first $m$ squares?
$endgroup$
– Arnaud D.
Aug 17 '18 at 9:55
$begingroup$
yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
$endgroup$
– user3243499
Aug 17 '18 at 10:00
$begingroup$
It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
$endgroup$
– Claude Leibovici
Aug 17 '18 at 10:03
$begingroup$
Where exactly does it stop?
$endgroup$
– Arnaud Mortier
Aug 17 '18 at 10:03
$begingroup$
@user3243499: no, it's the sum of the squares of the same parity.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:04
|
show 3 more comments
$begingroup$
Do you know the formula for the sum of the first $m$ squares?
$endgroup$
– Arnaud D.
Aug 17 '18 at 9:55
$begingroup$
yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
$endgroup$
– user3243499
Aug 17 '18 at 10:00
$begingroup$
It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
$endgroup$
– Claude Leibovici
Aug 17 '18 at 10:03
$begingroup$
Where exactly does it stop?
$endgroup$
– Arnaud Mortier
Aug 17 '18 at 10:03
$begingroup$
@user3243499: no, it's the sum of the squares of the same parity.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:04
$begingroup$
Do you know the formula for the sum of the first $m$ squares?
$endgroup$
– Arnaud D.
Aug 17 '18 at 9:55
$begingroup$
Do you know the formula for the sum of the first $m$ squares?
$endgroup$
– Arnaud D.
Aug 17 '18 at 9:55
$begingroup$
yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
$endgroup$
– user3243499
Aug 17 '18 at 10:00
$begingroup$
yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
$endgroup$
– user3243499
Aug 17 '18 at 10:00
$begingroup$
It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
$endgroup$
– Claude Leibovici
Aug 17 '18 at 10:03
$begingroup$
It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
$endgroup$
– Claude Leibovici
Aug 17 '18 at 10:03
$begingroup$
Where exactly does it stop?
$endgroup$
– Arnaud Mortier
Aug 17 '18 at 10:03
$begingroup$
Where exactly does it stop?
$endgroup$
– Arnaud Mortier
Aug 17 '18 at 10:03
$begingroup$
@user3243499: no, it's the sum of the squares of the same parity.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:04
$begingroup$
@user3243499: no, it's the sum of the squares of the same parity.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:04
|
show 3 more comments
5 Answers
5
active
oldest
votes
$begingroup$
It is well known that the sum of terms that are a polynomial expression of degree $d$, evaluated at the integers, is a polynomial expression of degree $d+1$.
So if you consider four values of the sum, you can obtain the requested expression as the Lagrangian interpolation polynomial on these four points.
For even $n$,
$$(0,0),(2,4),(4,20),(6,56)tofrac{n^3+3n^2+2n}6.$$
For odd $n$,
$$(1,1),(3,10),(5,35),(7,84)tofrac{n^3+3n^2+2n}6.$$
answered Aug 17 '18 at 10:11
Yves DaoustYves Daoust
129k675227
$endgroup$
add a comment |
$begingroup$
If $n$ is even, you have
$
begin{align}
sum_{k=1}^{frac{n}2}(2k)^2&=4sum_{k=1}^{frac{n}2}k^2\
&=4cdotfrac{frac{n}2(frac{n}2+1)(2cdotfrac{n}2+1)}6\
&=frac{n(n+1)(n+2)}6 \
text{If $n$ is odd, you have} \
\
sum_{k=1}^{frac{n+1}2}(2k-1)^2&=4sum_{k=1}^{frac{n+1}2}k^2-4sum_{k=1}^{frac{n+1}2}k+sum_{k=1}^{frac{n+1}2}1 \
&=4cdotfrac{frac{n+1}2(frac{n+1}2+1)(2cdotfrac{n+1}2+1)}6-4cdotfrac{frac{n+1}2(frac{n+1}2+1)}2+frac{n+1}2 \
&=frac{n(n+1)(n+2)}6
end{align}$
answered Aug 17 '18 at 11:53
Abraham ZhangAbraham Zhang
596312
$endgroup$
add a comment |
$begingroup$
Use the classic formula for the sum of increasing squares and some manipulation to find:
$$frac{1}{3} left(leftlfloor frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor
frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor frac{n-1}{2}rightrfloor
+3right)$$
answered Aug 17 '18 at 10:04
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
I don't think that the formula is right for the even $n$.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:17
add a comment |
$begingroup$
An alternative method:
Let's call the sum of alternate squares up to $n$ by $s(n)$.iafor $n=0,1,2,3,4,5,6$ is
$0,1,4,10,20,35, 56$
The differences between these terms are:
$1,3,6,10,15,21$
which are the triangular numbers $frac{n(n+1)}{2}$. So this suggests that
$s(n) = sum_{m=0}^{n} frac{m(m+1)}{2}$
$= sum_{m=0}^{n} (frac{m^2}{2} + frac{m}{2})$
$=frac{n(n+1)(2n+1)}{12} + frac{n(n+1)}{4}$
$=frac{n(n+1)(2n+1) + 3n(n+1)}{12}$
$ = frac{n(n+1)(n+2)}{6}$
This is not a formal proof, but it does suggest a formula which can form the starting point of a proof by induction
answered Aug 17 '18 at 11:29
gandalf61gandalf61
8,816725
$endgroup$
add a comment |
$begingroup$
Since $x^2=frac{x(x+1)}2+frac{(x-1)x}2$, your sum becomes $$sum_{k=1}^nfrac{k(k+1)}2=frac{n(n+1)(n+2)}6$$
answered Dec 28 '18 at 0:29
Abraham ZhangAbraham Zhang
596312
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is well known that the sum of terms that are a polynomial expression of degree $d$, evaluated at the integers, is a polynomial expression of degree $d+1$.
So if you consider four values of the sum, you can obtain the requested expression as the Lagrangian interpolation polynomial on these four points.
For even $n$,
$$(0,0),(2,4),(4,20),(6,56)tofrac{n^3+3n^2+2n}6.$$
For odd $n$,
$$(1,1),(3,10),(5,35),(7,84)tofrac{n^3+3n^2+2n}6.$$
answered Aug 17 '18 at 10:11
Yves DaoustYves Daoust
129k675227
$endgroup$
add a comment |
$begingroup$
It is well known that the sum of terms that are a polynomial expression of degree $d$, evaluated at the integers, is a polynomial expression of degree $d+1$.
So if you consider four values of the sum, you can obtain the requested expression as the Lagrangian interpolation polynomial on these four points.
For even $n$,
$$(0,0),(2,4),(4,20),(6,56)tofrac{n^3+3n^2+2n}6.$$
For odd $n$,
$$(1,1),(3,10),(5,35),(7,84)tofrac{n^3+3n^2+2n}6.$$
answered Aug 17 '18 at 10:11
Yves DaoustYves Daoust
129k675227
$endgroup$
add a comment |
$begingroup$
It is well known that the sum of terms that are a polynomial expression of degree $d$, evaluated at the integers, is a polynomial expression of degree $d+1$.
So if you consider four values of the sum, you can obtain the requested expression as the Lagrangian interpolation polynomial on these four points.
For even $n$,
$$(0,0),(2,4),(4,20),(6,56)tofrac{n^3+3n^2+2n}6.$$
For odd $n$,
$$(1,1),(3,10),(5,35),(7,84)tofrac{n^3+3n^2+2n}6.$$
answered Aug 17 '18 at 10:11
Yves DaoustYves Daoust
129k675227
$endgroup$
It is well known that the sum of terms that are a polynomial expression of degree $d$, evaluated at the integers, is a polynomial expression of degree $d+1$.
So if you consider four values of the sum, you can obtain the requested expression as the Lagrangian interpolation polynomial on these four points.
For even $n$,
$$(0,0),(2,4),(4,20),(6,56)tofrac{n^3+3n^2+2n}6.$$
For odd $n$,
$$(1,1),(3,10),(5,35),(7,84)tofrac{n^3+3n^2+2n}6.$$
answered Aug 17 '18 at 10:11
Yves DaoustYves Daoust
129k675227
answered Aug 17 '18 at 10:11
Yves DaoustYves Daoust
129k675227
answered Aug 17 '18 at 10:11
Yves DaoustYves Daoust
129k675227
answered Aug 17 '18 at 10:11
Yves DaoustYves Daoust
129k675227
129k675227
add a comment |
add a comment |
$begingroup$
If $n$ is even, you have
$
begin{align}
sum_{k=1}^{frac{n}2}(2k)^2&=4sum_{k=1}^{frac{n}2}k^2\
&=4cdotfrac{frac{n}2(frac{n}2+1)(2cdotfrac{n}2+1)}6\
&=frac{n(n+1)(n+2)}6 \
text{If $n$ is odd, you have} \
\
sum_{k=1}^{frac{n+1}2}(2k-1)^2&=4sum_{k=1}^{frac{n+1}2}k^2-4sum_{k=1}^{frac{n+1}2}k+sum_{k=1}^{frac{n+1}2}1 \
&=4cdotfrac{frac{n+1}2(frac{n+1}2+1)(2cdotfrac{n+1}2+1)}6-4cdotfrac{frac{n+1}2(frac{n+1}2+1)}2+frac{n+1}2 \
&=frac{n(n+1)(n+2)}6
end{align}$
answered Aug 17 '18 at 11:53
Abraham ZhangAbraham Zhang
596312
$endgroup$
add a comment |
$begingroup$
If $n$ is even, you have
$
begin{align}
sum_{k=1}^{frac{n}2}(2k)^2&=4sum_{k=1}^{frac{n}2}k^2\
&=4cdotfrac{frac{n}2(frac{n}2+1)(2cdotfrac{n}2+1)}6\
&=frac{n(n+1)(n+2)}6 \
text{If $n$ is odd, you have} \
\
sum_{k=1}^{frac{n+1}2}(2k-1)^2&=4sum_{k=1}^{frac{n+1}2}k^2-4sum_{k=1}^{frac{n+1}2}k+sum_{k=1}^{frac{n+1}2}1 \
&=4cdotfrac{frac{n+1}2(frac{n+1}2+1)(2cdotfrac{n+1}2+1)}6-4cdotfrac{frac{n+1}2(frac{n+1}2+1)}2+frac{n+1}2 \
&=frac{n(n+1)(n+2)}6
end{align}$
answered Aug 17 '18 at 11:53
Abraham ZhangAbraham Zhang
596312
$endgroup$
add a comment |
$begingroup$
If $n$ is even, you have
$
begin{align}
sum_{k=1}^{frac{n}2}(2k)^2&=4sum_{k=1}^{frac{n}2}k^2\
&=4cdotfrac{frac{n}2(frac{n}2+1)(2cdotfrac{n}2+1)}6\
&=frac{n(n+1)(n+2)}6 \
text{If $n$ is odd, you have} \
\
sum_{k=1}^{frac{n+1}2}(2k-1)^2&=4sum_{k=1}^{frac{n+1}2}k^2-4sum_{k=1}^{frac{n+1}2}k+sum_{k=1}^{frac{n+1}2}1 \
&=4cdotfrac{frac{n+1}2(frac{n+1}2+1)(2cdotfrac{n+1}2+1)}6-4cdotfrac{frac{n+1}2(frac{n+1}2+1)}2+frac{n+1}2 \
&=frac{n(n+1)(n+2)}6
end{align}$
answered Aug 17 '18 at 11:53
Abraham ZhangAbraham Zhang
596312
$endgroup$
If $n$ is even, you have
$
begin{align}
sum_{k=1}^{frac{n}2}(2k)^2&=4sum_{k=1}^{frac{n}2}k^2\
&=4cdotfrac{frac{n}2(frac{n}2+1)(2cdotfrac{n}2+1)}6\
&=frac{n(n+1)(n+2)}6 \
text{If $n$ is odd, you have} \
\
sum_{k=1}^{frac{n+1}2}(2k-1)^2&=4sum_{k=1}^{frac{n+1}2}k^2-4sum_{k=1}^{frac{n+1}2}k+sum_{k=1}^{frac{n+1}2}1 \
&=4cdotfrac{frac{n+1}2(frac{n+1}2+1)(2cdotfrac{n+1}2+1)}6-4cdotfrac{frac{n+1}2(frac{n+1}2+1)}2+frac{n+1}2 \
&=frac{n(n+1)(n+2)}6
end{align}$
answered Aug 17 '18 at 11:53
Abraham ZhangAbraham Zhang
596312
answered Aug 17 '18 at 11:53
Abraham ZhangAbraham Zhang
596312
answered Aug 17 '18 at 11:53
Abraham ZhangAbraham Zhang
596312
answered Aug 17 '18 at 11:53
Abraham ZhangAbraham Zhang
596312
596312
add a comment |
add a comment |
$begingroup$
Use the classic formula for the sum of increasing squares and some manipulation to find:
$$frac{1}{3} left(leftlfloor frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor
frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor frac{n-1}{2}rightrfloor
+3right)$$
answered Aug 17 '18 at 10:04
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
I don't think that the formula is right for the even $n$.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:17
add a comment |
$begingroup$
Use the classic formula for the sum of increasing squares and some manipulation to find:
$$frac{1}{3} left(leftlfloor frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor
frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor frac{n-1}{2}rightrfloor
+3right)$$
answered Aug 17 '18 at 10:04
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
I don't think that the formula is right for the even $n$.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:17
add a comment |
$begingroup$
Use the classic formula for the sum of increasing squares and some manipulation to find:
$$frac{1}{3} left(leftlfloor frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor
frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor frac{n-1}{2}rightrfloor
+3right)$$
answered Aug 17 '18 at 10:04
David G. StorkDavid G. Stork
11k41432
$endgroup$
Use the classic formula for the sum of increasing squares and some manipulation to find:
$$frac{1}{3} left(leftlfloor frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor
frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor frac{n-1}{2}rightrfloor
+3right)$$
answered Aug 17 '18 at 10:04
David G. StorkDavid G. Stork
11k41432
answered Aug 17 '18 at 10:04
David G. StorkDavid G. Stork
11k41432
answered Aug 17 '18 at 10:04
David G. StorkDavid G. Stork
11k41432
answered Aug 17 '18 at 10:04
David G. StorkDavid G. Stork
11k41432
11k41432
$begingroup$
I don't think that the formula is right for the even $n$.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:17
add a comment |
$begingroup$
I don't think that the formula is right for the even $n$.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:17
$begingroup$
I don't think that the formula is right for the even $n$.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:17
$begingroup$
I don't think that the formula is right for the even $n$.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:17
add a comment |
$begingroup$
An alternative method:
Let's call the sum of alternate squares up to $n$ by $s(n)$.iafor $n=0,1,2,3,4,5,6$ is
$0,1,4,10,20,35, 56$
The differences between these terms are:
$1,3,6,10,15,21$
which are the triangular numbers $frac{n(n+1)}{2}$. So this suggests that
$s(n) = sum_{m=0}^{n} frac{m(m+1)}{2}$
$= sum_{m=0}^{n} (frac{m^2}{2} + frac{m}{2})$
$=frac{n(n+1)(2n+1)}{12} + frac{n(n+1)}{4}$
$=frac{n(n+1)(2n+1) + 3n(n+1)}{12}$
$ = frac{n(n+1)(n+2)}{6}$
This is not a formal proof, but it does suggest a formula which can form the starting point of a proof by induction
answered Aug 17 '18 at 11:29
gandalf61gandalf61
8,816725
$endgroup$
add a comment |
$begingroup$
An alternative method:
Let's call the sum of alternate squares up to $n$ by $s(n)$.iafor $n=0,1,2,3,4,5,6$ is
$0,1,4,10,20,35, 56$
The differences between these terms are:
$1,3,6,10,15,21$
which are the triangular numbers $frac{n(n+1)}{2}$. So this suggests that
$s(n) = sum_{m=0}^{n} frac{m(m+1)}{2}$
$= sum_{m=0}^{n} (frac{m^2}{2} + frac{m}{2})$
$=frac{n(n+1)(2n+1)}{12} + frac{n(n+1)}{4}$
$=frac{n(n+1)(2n+1) + 3n(n+1)}{12}$
$ = frac{n(n+1)(n+2)}{6}$
This is not a formal proof, but it does suggest a formula which can form the starting point of a proof by induction
answered Aug 17 '18 at 11:29
gandalf61gandalf61
8,816725
$endgroup$
add a comment |
$begingroup$
An alternative method:
Let's call the sum of alternate squares up to $n$ by $s(n)$.iafor $n=0,1,2,3,4,5,6$ is
$0,1,4,10,20,35, 56$
The differences between these terms are:
$1,3,6,10,15,21$
which are the triangular numbers $frac{n(n+1)}{2}$. So this suggests that
$s(n) = sum_{m=0}^{n} frac{m(m+1)}{2}$
$= sum_{m=0}^{n} (frac{m^2}{2} + frac{m}{2})$
$=frac{n(n+1)(2n+1)}{12} + frac{n(n+1)}{4}$
$=frac{n(n+1)(2n+1) + 3n(n+1)}{12}$
$ = frac{n(n+1)(n+2)}{6}$
This is not a formal proof, but it does suggest a formula which can form the starting point of a proof by induction
answered Aug 17 '18 at 11:29
gandalf61gandalf61
8,816725
$endgroup$
An alternative method:
Let's call the sum of alternate squares up to $n$ by $s(n)$.iafor $n=0,1,2,3,4,5,6$ is
$0,1,4,10,20,35, 56$
The differences between these terms are:
$1,3,6,10,15,21$
which are the triangular numbers $frac{n(n+1)}{2}$. So this suggests that
$s(n) = sum_{m=0}^{n} frac{m(m+1)}{2}$
$= sum_{m=0}^{n} (frac{m^2}{2} + frac{m}{2})$
$=frac{n(n+1)(2n+1)}{12} + frac{n(n+1)}{4}$
$=frac{n(n+1)(2n+1) + 3n(n+1)}{12}$
$ = frac{n(n+1)(n+2)}{6}$
This is not a formal proof, but it does suggest a formula which can form the starting point of a proof by induction
answered Aug 17 '18 at 11:29
gandalf61gandalf61
8,816725
answered Aug 17 '18 at 11:29
gandalf61gandalf61
8,816725
answered Aug 17 '18 at 11:29
gandalf61gandalf61
8,816725
answered Aug 17 '18 at 11:29
gandalf61gandalf61
8,816725
8,816725
add a comment |
add a comment |
$begingroup$
Since $x^2=frac{x(x+1)}2+frac{(x-1)x}2$, your sum becomes $$sum_{k=1}^nfrac{k(k+1)}2=frac{n(n+1)(n+2)}6$$
answered Dec 28 '18 at 0:29
Abraham ZhangAbraham Zhang
596312
$endgroup$
add a comment |
$begingroup$
Since $x^2=frac{x(x+1)}2+frac{(x-1)x}2$, your sum becomes $$sum_{k=1}^nfrac{k(k+1)}2=frac{n(n+1)(n+2)}6$$
answered Dec 28 '18 at 0:29
Abraham ZhangAbraham Zhang
596312
$endgroup$
add a comment |
$begingroup$
Since $x^2=frac{x(x+1)}2+frac{(x-1)x}2$, your sum becomes $$sum_{k=1}^nfrac{k(k+1)}2=frac{n(n+1)(n+2)}6$$
answered Dec 28 '18 at 0:29
Abraham ZhangAbraham Zhang
596312
$endgroup$
Since $x^2=frac{x(x+1)}2+frac{(x-1)x}2$, your sum becomes $$sum_{k=1}^nfrac{k(k+1)}2=frac{n(n+1)(n+2)}6$$
answered Dec 28 '18 at 0:29
Abraham ZhangAbraham Zhang
596312
answered Dec 28 '18 at 0:29
Abraham ZhangAbraham Zhang
596312
answered Dec 28 '18 at 0:29
Abraham ZhangAbraham Zhang
596312
answered Dec 28 '18 at 0:29
Abraham ZhangAbraham Zhang
596312
596312
add a comment |
add a comment |
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$begingroup$
Do you know the formula for the sum of the first $m$ squares?
$endgroup$
– Arnaud D.
Aug 17 '18 at 9:55
$begingroup$
yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
$endgroup$
– user3243499
Aug 17 '18 at 10:00
$begingroup$
It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
$endgroup$
– Claude Leibovici
Aug 17 '18 at 10:03
$begingroup$
Where exactly does it stop?
$endgroup$
– Arnaud Mortier
Aug 17 '18 at 10:03
$begingroup$
@user3243499: no, it's the sum of the squares of the same parity.
$endgroup$
– Yves Daoust
Aug 17 '18 at 10:04