Formula for the series $1+1+4+1+4+9..$ [on hold]
up vote
-3
down vote
favorite
Derive a formula for the series
$f(n)=1+1+4+1+4+9+.....+1+4+9+16...n^2$
Example
$f(3)= 1+1+4+1+4+9$
I couldn't really find any derivation of this online or a question like this it self.
sequences-and-series
edited Nov 25 at 16:57
amWhy
191k27223438
asked Nov 25 at 15:26
Ktk
13
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Ktk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by José Carlos Santos, TheSimpliFire, kingW3, amWhy, Did Nov 25 at 18:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, TheSimpliFire, kingW3, amWhy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-3
down vote
favorite
Derive a formula for the series
$f(n)=1+1+4+1+4+9+.....+1+4+9+16...n^2$
Example
$f(3)= 1+1+4+1+4+9$
I couldn't really find any derivation of this online or a question like this it self.
sequences-and-series
edited Nov 25 at 16:57
amWhy
191k27223438
asked Nov 25 at 15:26
Ktk
13
New contributor
Ktk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by José Carlos Santos, TheSimpliFire, kingW3, amWhy, Did Nov 25 at 18:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, TheSimpliFire, kingW3, amWhy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
1
First thing is to figure out what is $1+4+cdots+n^2$
– kingW3
Nov 25 at 15:29
$f(n)=displaystylesum_{k=0}^n dfrac{k(k+1)(2k+1)}{6}$
– Yadati Kiran
Nov 25 at 15:30
@YadatiKiran: that is correct for kingW3's hint, but not for the original problem where $f(n)$ is defined
– Ross Millikan
Nov 25 at 15:31
You could look up Faulhaber's formula
– Ross Millikan
Nov 25 at 15:32
1
@RossMillikan: Sum of squares is $dfrac{k(k+1)(2k+1)}{6}$ and here $f(n)=1+(1+4)+(1+4+9)+cdots+(1+4+9+cdots+n^2)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}$. So it is that sum i suppose.
– Yadati Kiran
Nov 25 at 15:35
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
Derive a formula for the series
$f(n)=1+1+4+1+4+9+.....+1+4+9+16...n^2$
Example
$f(3)= 1+1+4+1+4+9$
I couldn't really find any derivation of this online or a question like this it self.
sequences-and-series
edited Nov 25 at 16:57
amWhy
191k27223438
asked Nov 25 at 15:26
Ktk
13
New contributor
Ktk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Derive a formula for the series
$f(n)=1+1+4+1+4+9+.....+1+4+9+16...n^2$
Example
$f(3)= 1+1+4+1+4+9$
I couldn't really find any derivation of this online or a question like this it self.
sequences-and-series
sequences-and-series
edited Nov 25 at 16:57
amWhy
191k27223438
asked Nov 25 at 15:26
Ktk
13
New contributor
Ktk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 25 at 16:57
amWhy
191k27223438
asked Nov 25 at 15:26
Ktk
13
New contributor
Ktk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 25 at 16:57
amWhy
191k27223438
edited Nov 25 at 16:57
amWhy
191k27223438
edited Nov 25 at 16:57
amWhy
191k27223438
191k27223438
asked Nov 25 at 15:26
Ktk
13
New contributor
Ktk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 25 at 15:26
Ktk
13
asked Nov 25 at 15:26
Ktk
13
13
New contributor
Ktk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ktk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ktk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by José Carlos Santos, TheSimpliFire, kingW3, amWhy, Did Nov 25 at 18:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, TheSimpliFire, kingW3, amWhy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by José Carlos Santos, TheSimpliFire, kingW3, amWhy, Did Nov 25 at 18:27
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, TheSimpliFire, kingW3, amWhy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
1
First thing is to figure out what is $1+4+cdots+n^2$
– kingW3
Nov 25 at 15:29
$f(n)=displaystylesum_{k=0}^n dfrac{k(k+1)(2k+1)}{6}$
– Yadati Kiran
Nov 25 at 15:30
@YadatiKiran: that is correct for kingW3's hint, but not for the original problem where $f(n)$ is defined
– Ross Millikan
Nov 25 at 15:31
You could look up Faulhaber's formula
– Ross Millikan
Nov 25 at 15:32
1
@RossMillikan: Sum of squares is $dfrac{k(k+1)(2k+1)}{6}$ and here $f(n)=1+(1+4)+(1+4+9)+cdots+(1+4+9+cdots+n^2)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}$. So it is that sum i suppose.
– Yadati Kiran
Nov 25 at 15:35
add a comment |
1
First thing is to figure out what is $1+4+cdots+n^2$
– kingW3
Nov 25 at 15:29
$f(n)=displaystylesum_{k=0}^n dfrac{k(k+1)(2k+1)}{6}$
– Yadati Kiran
Nov 25 at 15:30
@YadatiKiran: that is correct for kingW3's hint, but not for the original problem where $f(n)$ is defined
– Ross Millikan
Nov 25 at 15:31
You could look up Faulhaber's formula
– Ross Millikan
Nov 25 at 15:32
1
@RossMillikan: Sum of squares is $dfrac{k(k+1)(2k+1)}{6}$ and here $f(n)=1+(1+4)+(1+4+9)+cdots+(1+4+9+cdots+n^2)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}$. So it is that sum i suppose.
– Yadati Kiran
Nov 25 at 15:35
1
1
First thing is to figure out what is $1+4+cdots+n^2$
– kingW3
Nov 25 at 15:29
First thing is to figure out what is $1+4+cdots+n^2$
– kingW3
Nov 25 at 15:29
$f(n)=displaystylesum_{k=0}^n dfrac{k(k+1)(2k+1)}{6}$
– Yadati Kiran
Nov 25 at 15:30
$f(n)=displaystylesum_{k=0}^n dfrac{k(k+1)(2k+1)}{6}$
– Yadati Kiran
Nov 25 at 15:30
@YadatiKiran: that is correct for kingW3's hint, but not for the original problem where $f(n)$ is defined
– Ross Millikan
Nov 25 at 15:31
@YadatiKiran: that is correct for kingW3's hint, but not for the original problem where $f(n)$ is defined
– Ross Millikan
Nov 25 at 15:31
You could look up Faulhaber's formula
– Ross Millikan
Nov 25 at 15:32
You could look up Faulhaber's formula
– Ross Millikan
Nov 25 at 15:32
1
1
@RossMillikan: Sum of squares is $dfrac{k(k+1)(2k+1)}{6}$ and here $f(n)=1+(1+4)+(1+4+9)+cdots+(1+4+9+cdots+n^2)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}$. So it is that sum i suppose.
– Yadati Kiran
Nov 25 at 15:35
@RossMillikan: Sum of squares is $dfrac{k(k+1)(2k+1)}{6}$ and here $f(n)=1+(1+4)+(1+4+9)+cdots+(1+4+9+cdots+n^2)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}$. So it is that sum i suppose.
– Yadati Kiran
Nov 25 at 15:35
add a comment |
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
$f(n)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}=sum_{k=1}^n dfrac{2k^3+3k^2+k}{6}=dfrac{1}{6}left(2sum_{k=1}^n k^3+3sum_{k=1}^nk^2+sum_{k=1}^nkright)$.
(Assuming you know the formula for the sum of cubes and squares of first $n$ natural numbers.)
answered Nov 25 at 15:43
Yadati Kiran
1,288317
add a comment |
up vote
2
down vote
As suggested by the user Yadati Kiran, $$f(n)=frac {1}{12} n(n+1)(n^2+3n+2)$$
answered Nov 25 at 15:43
Awe Kumar Jha
3189
add a comment |
up vote
0
down vote
Hint: This is just
$$sum_{k=1}^n left(sum_{j=1}^k j^2right)$$
answered Nov 25 at 15:52
MPW
29.6k11856
add a comment |
up vote
0
down vote
With application of hockeystick identity:
$$begin{aligned}sum_{k=1}^{n}sum_{m=1}^{k}m^{2} & =sum_{k=1}^{n}sum_{m=1}^{k}left[2binom{m}{2}+binom{m}{1}right]\
& =2sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{2}+sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{1}\
& =2sum_{k=1}^{n}binom{k+1}{3}+sum_{k=1}^{n}binom{k+1}{2}\
& =2binom{n+2}{4}+binom{n+2}{3}\
& =frac{1}{12}left(n+2right)left(n+1right)nleft(n-1right)+frac{1}{6}left(n+2right)left(n+1right)n\
& =frac{1}{12}left(n+2right)left(n+1right)^{2}n
end{aligned}
$$
answered Nov 25 at 16:10
drhab
94.8k543125
In second line first summation shouldn't $m=2$ to $k$ for applying hockey stick identity?
– Yadati Kiran
Nov 25 at 17:02
@YadatiKiran does not really matter. Use the (useful ) convention that $binom {r}{s}:=0$ if $snotin {0,1,dots,r} $.
– drhab
Nov 25 at 17:31
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$f(n)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}=sum_{k=1}^n dfrac{2k^3+3k^2+k}{6}=dfrac{1}{6}left(2sum_{k=1}^n k^3+3sum_{k=1}^nk^2+sum_{k=1}^nkright)$.
(Assuming you know the formula for the sum of cubes and squares of first $n$ natural numbers.)
answered Nov 25 at 15:43
Yadati Kiran
1,288317
add a comment |
up vote
3
down vote
accepted
$f(n)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}=sum_{k=1}^n dfrac{2k^3+3k^2+k}{6}=dfrac{1}{6}left(2sum_{k=1}^n k^3+3sum_{k=1}^nk^2+sum_{k=1}^nkright)$.
(Assuming you know the formula for the sum of cubes and squares of first $n$ natural numbers.)
answered Nov 25 at 15:43
Yadati Kiran
1,288317
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$f(n)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}=sum_{k=1}^n dfrac{2k^3+3k^2+k}{6}=dfrac{1}{6}left(2sum_{k=1}^n k^3+3sum_{k=1}^nk^2+sum_{k=1}^nkright)$.
(Assuming you know the formula for the sum of cubes and squares of first $n$ natural numbers.)
answered Nov 25 at 15:43
Yadati Kiran
1,288317
$f(n)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}=sum_{k=1}^n dfrac{2k^3+3k^2+k}{6}=dfrac{1}{6}left(2sum_{k=1}^n k^3+3sum_{k=1}^nk^2+sum_{k=1}^nkright)$.
(Assuming you know the formula for the sum of cubes and squares of first $n$ natural numbers.)
answered Nov 25 at 15:43
Yadati Kiran
1,288317
answered Nov 25 at 15:43
Yadati Kiran
1,288317
answered Nov 25 at 15:43
Yadati Kiran
1,288317
answered Nov 25 at 15:43
Yadati Kiran
1,288317
1,288317
add a comment |
add a comment |
up vote
2
down vote
As suggested by the user Yadati Kiran, $$f(n)=frac {1}{12} n(n+1)(n^2+3n+2)$$
answered Nov 25 at 15:43
Awe Kumar Jha
3189
add a comment |
up vote
2
down vote
As suggested by the user Yadati Kiran, $$f(n)=frac {1}{12} n(n+1)(n^2+3n+2)$$
answered Nov 25 at 15:43
Awe Kumar Jha
3189
add a comment |
up vote
2
down vote
up vote
2
down vote
As suggested by the user Yadati Kiran, $$f(n)=frac {1}{12} n(n+1)(n^2+3n+2)$$
answered Nov 25 at 15:43
Awe Kumar Jha
3189
As suggested by the user Yadati Kiran, $$f(n)=frac {1}{12} n(n+1)(n^2+3n+2)$$
answered Nov 25 at 15:43
Awe Kumar Jha
3189
answered Nov 25 at 15:43
Awe Kumar Jha
3189
answered Nov 25 at 15:43
Awe Kumar Jha
3189
answered Nov 25 at 15:43
Awe Kumar Jha
3189
3189
add a comment |
add a comment |
up vote
0
down vote
Hint: This is just
$$sum_{k=1}^n left(sum_{j=1}^k j^2right)$$
answered Nov 25 at 15:52
MPW
29.6k11856
add a comment |
up vote
0
down vote
Hint: This is just
$$sum_{k=1}^n left(sum_{j=1}^k j^2right)$$
answered Nov 25 at 15:52
MPW
29.6k11856
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: This is just
$$sum_{k=1}^n left(sum_{j=1}^k j^2right)$$
answered Nov 25 at 15:52
MPW
29.6k11856
Hint: This is just
$$sum_{k=1}^n left(sum_{j=1}^k j^2right)$$
answered Nov 25 at 15:52
MPW
29.6k11856
answered Nov 25 at 15:52
MPW
29.6k11856
answered Nov 25 at 15:52
MPW
29.6k11856
answered Nov 25 at 15:52
MPW
29.6k11856
29.6k11856
add a comment |
add a comment |
up vote
0
down vote
With application of hockeystick identity:
$$begin{aligned}sum_{k=1}^{n}sum_{m=1}^{k}m^{2} & =sum_{k=1}^{n}sum_{m=1}^{k}left[2binom{m}{2}+binom{m}{1}right]\
& =2sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{2}+sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{1}\
& =2sum_{k=1}^{n}binom{k+1}{3}+sum_{k=1}^{n}binom{k+1}{2}\
& =2binom{n+2}{4}+binom{n+2}{3}\
& =frac{1}{12}left(n+2right)left(n+1right)nleft(n-1right)+frac{1}{6}left(n+2right)left(n+1right)n\
& =frac{1}{12}left(n+2right)left(n+1right)^{2}n
end{aligned}
$$
answered Nov 25 at 16:10
drhab
94.8k543125
In second line first summation shouldn't $m=2$ to $k$ for applying hockey stick identity?
– Yadati Kiran
Nov 25 at 17:02
@YadatiKiran does not really matter. Use the (useful ) convention that $binom {r}{s}:=0$ if $snotin {0,1,dots,r} $.
– drhab
Nov 25 at 17:31
add a comment |
up vote
0
down vote
With application of hockeystick identity:
$$begin{aligned}sum_{k=1}^{n}sum_{m=1}^{k}m^{2} & =sum_{k=1}^{n}sum_{m=1}^{k}left[2binom{m}{2}+binom{m}{1}right]\
& =2sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{2}+sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{1}\
& =2sum_{k=1}^{n}binom{k+1}{3}+sum_{k=1}^{n}binom{k+1}{2}\
& =2binom{n+2}{4}+binom{n+2}{3}\
& =frac{1}{12}left(n+2right)left(n+1right)nleft(n-1right)+frac{1}{6}left(n+2right)left(n+1right)n\
& =frac{1}{12}left(n+2right)left(n+1right)^{2}n
end{aligned}
$$
answered Nov 25 at 16:10
drhab
94.8k543125
In second line first summation shouldn't $m=2$ to $k$ for applying hockey stick identity?
– Yadati Kiran
Nov 25 at 17:02
@YadatiKiran does not really matter. Use the (useful ) convention that $binom {r}{s}:=0$ if $snotin {0,1,dots,r} $.
– drhab
Nov 25 at 17:31
add a comment |
up vote
0
down vote
up vote
0
down vote
With application of hockeystick identity:
$$begin{aligned}sum_{k=1}^{n}sum_{m=1}^{k}m^{2} & =sum_{k=1}^{n}sum_{m=1}^{k}left[2binom{m}{2}+binom{m}{1}right]\
& =2sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{2}+sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{1}\
& =2sum_{k=1}^{n}binom{k+1}{3}+sum_{k=1}^{n}binom{k+1}{2}\
& =2binom{n+2}{4}+binom{n+2}{3}\
& =frac{1}{12}left(n+2right)left(n+1right)nleft(n-1right)+frac{1}{6}left(n+2right)left(n+1right)n\
& =frac{1}{12}left(n+2right)left(n+1right)^{2}n
end{aligned}
$$
answered Nov 25 at 16:10
drhab
94.8k543125
With application of hockeystick identity:
$$begin{aligned}sum_{k=1}^{n}sum_{m=1}^{k}m^{2} & =sum_{k=1}^{n}sum_{m=1}^{k}left[2binom{m}{2}+binom{m}{1}right]\
& =2sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{2}+sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{1}\
& =2sum_{k=1}^{n}binom{k+1}{3}+sum_{k=1}^{n}binom{k+1}{2}\
& =2binom{n+2}{4}+binom{n+2}{3}\
& =frac{1}{12}left(n+2right)left(n+1right)nleft(n-1right)+frac{1}{6}left(n+2right)left(n+1right)n\
& =frac{1}{12}left(n+2right)left(n+1right)^{2}n
end{aligned}
$$
answered Nov 25 at 16:10
drhab
94.8k543125
answered Nov 25 at 16:10
drhab
94.8k543125
answered Nov 25 at 16:10
drhab
94.8k543125
answered Nov 25 at 16:10
drhab
94.8k543125
94.8k543125
In second line first summation shouldn't $m=2$ to $k$ for applying hockey stick identity?
– Yadati Kiran
Nov 25 at 17:02
@YadatiKiran does not really matter. Use the (useful ) convention that $binom {r}{s}:=0$ if $snotin {0,1,dots,r} $.
– drhab
Nov 25 at 17:31
add a comment |
In second line first summation shouldn't $m=2$ to $k$ for applying hockey stick identity?
– Yadati Kiran
Nov 25 at 17:02
@YadatiKiran does not really matter. Use the (useful ) convention that $binom {r}{s}:=0$ if $snotin {0,1,dots,r} $.
– drhab
Nov 25 at 17:31
In second line first summation shouldn't $m=2$ to $k$ for applying hockey stick identity?
– Yadati Kiran
Nov 25 at 17:02
In second line first summation shouldn't $m=2$ to $k$ for applying hockey stick identity?
– Yadati Kiran
Nov 25 at 17:02
@YadatiKiran does not really matter. Use the (useful ) convention that $binom {r}{s}:=0$ if $snotin {0,1,dots,r} $.
– drhab
Nov 25 at 17:31
@YadatiKiran does not really matter. Use the (useful ) convention that $binom {r}{s}:=0$ if $snotin {0,1,dots,r} $.
– drhab
Nov 25 at 17:31
add a comment |
1
First thing is to figure out what is $1+4+cdots+n^2$
– kingW3
Nov 25 at 15:29
$f(n)=displaystylesum_{k=0}^n dfrac{k(k+1)(2k+1)}{6}$
– Yadati Kiran
Nov 25 at 15:30
@YadatiKiran: that is correct for kingW3's hint, but not for the original problem where $f(n)$ is defined
– Ross Millikan
Nov 25 at 15:31
You could look up Faulhaber's formula
– Ross Millikan
Nov 25 at 15:32
1
@RossMillikan: Sum of squares is $dfrac{k(k+1)(2k+1)}{6}$ and here $f(n)=1+(1+4)+(1+4+9)+cdots+(1+4+9+cdots+n^2)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}$. So it is that sum i suppose.
– Yadati Kiran
Nov 25 at 15:35