Mixing
Sum of a Sequence of Odd Numbers that are Squared [duplicate]
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This question already has an answer here:
Calculate sum of squares of first n odd numbers
5 answers
What is the sum of all the numbers in the sequence $1^2 + 3^2 + 5^2 + 7^2 + 9^2 + ldots + k^2$. Note that all the numbers being squared in the sequence are all odd numbers.
This is what I have done so far (sorry if the images are an inconvenience, but this was the clearest way to display my working out):
I am a little stuck on what to do next and how to obtain $fracn (4n^2 - 1)3$ as the final result as this is what I am meant to end up with. It would be really appreciated if anyone could make suggestions towards completing and improving my method. Thanks! :)
discrete-mathematics summation sums-of-squares telescopic-series
edited 2 days ago
Michael Rozenberg
86.9k1576178
asked 2 days ago
Cameron Choi
474
marked as duplicate by Jyrki Lahtonen, amWhy discrete-mathematics
Users with the  discrete-mathematics badge can single-handedly close discrete-mathematics questions as duplicates and reopen them as needed.
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2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
 |Â
show 6 more comments
up vote
3
down vote
favorite
This question already has an answer here:
Calculate sum of squares of first n odd numbers
5 answers
What is the sum of all the numbers in the sequence $1^2 + 3^2 + 5^2 + 7^2 + 9^2 + ldots + k^2$. Note that all the numbers being squared in the sequence are all odd numbers.
This is what I have done so far (sorry if the images are an inconvenience, but this was the clearest way to display my working out):
I am a little stuck on what to do next and how to obtain $fracn (4n^2 - 1)3$ as the final result as this is what I am meant to end up with. It would be really appreciated if anyone could make suggestions towards completing and improving my method. Thanks! :)
discrete-mathematics summation sums-of-squares telescopic-series
edited 2 days ago
Michael Rozenberg
86.9k1576178
asked 2 days ago
Cameron Choi
474
marked as duplicate by Jyrki Lahtonen, amWhy discrete-mathematics
Users with the  discrete-mathematics badge can single-handedly close discrete-mathematics questions as duplicates and reopen them as needed.
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2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Do you know proof by induction?
– rbird
2 days ago
I only know it very vaguely, sorry.
– Cameron Choi
2 days ago
1
Downvoting all "trusted" users who answer an obvious dupe.
– Jyrki Lahtonen
2 days ago
1
@Jyrki Lahtonen OP asked for a verification of the proof. I corrected his/her errors in bold font.
– Robert Z
2 days ago
@RobertZ You may disagree but I am not convinced by the excuse that because this user made a different error from the previous asker we should keep ten versions of an elementary calculation.
– Jyrki Lahtonen
2 days ago
 |Â
show 6 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This question already has an answer here:
Calculate sum of squares of first n odd numbers
5 answers
What is the sum of all the numbers in the sequence $1^2 + 3^2 + 5^2 + 7^2 + 9^2 + ldots + k^2$. Note that all the numbers being squared in the sequence are all odd numbers.
This is what I have done so far (sorry if the images are an inconvenience, but this was the clearest way to display my working out):
I am a little stuck on what to do next and how to obtain $fracn (4n^2 - 1)3$ as the final result as this is what I am meant to end up with. It would be really appreciated if anyone could make suggestions towards completing and improving my method. Thanks! :)
discrete-mathematics summation sums-of-squares telescopic-series
edited 2 days ago
Michael Rozenberg
86.9k1576178
asked 2 days ago
Cameron Choi
474
This question already has an answer here:
Calculate sum of squares of first n odd numbers
5 answers
What is the sum of all the numbers in the sequence $1^2 + 3^2 + 5^2 + 7^2 + 9^2 + ldots + k^2$. Note that all the numbers being squared in the sequence are all odd numbers.
This is what I have done so far (sorry if the images are an inconvenience, but this was the clearest way to display my working out):
I am a little stuck on what to do next and how to obtain $fracn (4n^2 - 1)3$ as the final result as this is what I am meant to end up with. It would be really appreciated if anyone could make suggestions towards completing and improving my method. Thanks! :)
This question already has an answer here:
Calculate sum of squares of first n odd numbers
5 answers
discrete-mathematics summation sums-of-squares telescopic-series
edited 2 days ago
Michael Rozenberg
86.9k1576178
asked 2 days ago
Cameron Choi
474
edited 2 days ago
Michael Rozenberg
86.9k1576178
edited 2 days ago
Michael Rozenberg
86.9k1576178
edited 2 days ago
Michael Rozenberg
86.9k1576178
86.9k1576178
asked 2 days ago
Cameron Choi
474
asked 2 days ago
Cameron Choi
474
asked 2 days ago
Cameron Choi
474
474
marked as duplicate by Jyrki Lahtonen, amWhy discrete-mathematics
Users with the  discrete-mathematics badge can single-handedly close discrete-mathematics questions as duplicates and reopen them as needed.
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jyrki Lahtonen, amWhy discrete-mathematics
Users with the  discrete-mathematics badge can single-handedly close discrete-mathematics questions as duplicates and reopen them as needed.
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2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Do you know proof by induction?
– rbird
2 days ago
I only know it very vaguely, sorry.
– Cameron Choi
2 days ago
1
Downvoting all "trusted" users who answer an obvious dupe.
– Jyrki Lahtonen
2 days ago
1
@Jyrki Lahtonen OP asked for a verification of the proof. I corrected his/her errors in bold font.
– Robert Z
2 days ago
@RobertZ You may disagree but I am not convinced by the excuse that because this user made a different error from the previous asker we should keep ten versions of an elementary calculation.
– Jyrki Lahtonen
2 days ago
 |Â
show 6 more comments
Do you know proof by induction?
– rbird
2 days ago
I only know it very vaguely, sorry.
– Cameron Choi
2 days ago
1
Downvoting all "trusted" users who answer an obvious dupe.
– Jyrki Lahtonen
2 days ago
1
@Jyrki Lahtonen OP asked for a verification of the proof. I corrected his/her errors in bold font.
– Robert Z
2 days ago
@RobertZ You may disagree but I am not convinced by the excuse that because this user made a different error from the previous asker we should keep ten versions of an elementary calculation.
– Jyrki Lahtonen
2 days ago
Do you know proof by induction?
– rbird
2 days ago
Do you know proof by induction?
– rbird
2 days ago
I only know it very vaguely, sorry.
– Cameron Choi
2 days ago
I only know it very vaguely, sorry.
– Cameron Choi
2 days ago
1
1
Downvoting all "trusted" users who answer an obvious dupe.
– Jyrki Lahtonen
2 days ago
Downvoting all "trusted" users who answer an obvious dupe.
– Jyrki Lahtonen
2 days ago
1
1
@Jyrki Lahtonen OP asked for a verification of the proof. I corrected his/her errors in bold font.
– Robert Z
2 days ago
@Jyrki Lahtonen OP asked for a verification of the proof. I corrected his/her errors in bold font.
– Robert Z
2 days ago
@RobertZ You may disagree but I am not convinced by the excuse that because this user made a different error from the previous asker we should keep ten versions of an elementary calculation.
– Jyrki Lahtonen
2 days ago
@RobertZ You may disagree but I am not convinced by the excuse that because this user made a different error from the previous asker we should keep ten versions of an elementary calculation.
– Jyrki Lahtonen
2 days ago
 |Â
show 6 more comments
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Your approach is almost correct. Check again your steps. At the end you should have
$$beginalign(k+mathbf2)^3-1
&=6(1+3^2+dots+k^2)+12(1+3+dots+k)+underbrace(8+8+dots+8)_text$(k+1)/2$ times\
&=6(1+3^2+dots+k^2)+12left(frack+12right)^2+8mathbfleft(frack+12right).endalign$$
Hence
$$6(1+3^2+dots+k^2)=(k+mathbf2)^3-1-12left(frack+12right)^2-8mathbfleft(frack+12right)$$
and it follows that
$$sum_j=1^n(2j-1)^2=1+3^2+dots+k^2=frack(k+2)(k+1)6=fracn (4n^2 - 1)3$$
where $n=(k+1)/2$.
edited 2 days ago
Jyrki Lahtonen
104k12161355
answered 2 days ago
Robert Z
83.4k954122
If I was to find the same sum, but instead with even numbers, would I write: 6(1 + 3^2 + ⋯ + k^2) = (k+2)^3 − 2 − 12((k^2)/4 + k/2) − 8((k+1)/2)
– Cameron Choi
2 days ago
@CameronChoi It should be $6(2^2 + dots + k^2) = (k+2)^3 − 2^3 − 12(k^2/4 + k/2) − 8(k/2) $ where $n=k/2$.
– Robert Z
2 days ago
add a comment |Â
up vote
1
down vote
Set $n=2m$
$$(2m+2)^3-(2m)^3=24m^2+24m+8=6(2m+1)^2+2$$
$$impliessum_m=0^n(6(2m+1)^2+2)=sum_m=0^n((2m+2)^3-(2m)^3)=sum_m=0^n(f(m+1)-f(m))$$ where $f(m)=(2m+2)^3$
$$6sum_m=0^n(2m+1)^2+2sum_m=0^n1=f(n+1)-f(0)=?$$
answered 2 days ago
lab bhattacharjee
214k14152263
add a comment |Â
up vote
0
down vote
The telescopic sum helps very well:
$$sum_k=1^n(2k-1)^2=sum_k=1^n(4k^2-4k+1)=sum_k=1^nleft(frac43(k^3-(k-1)^3)-frac13right)=$$
$$=frac43(n^3-0)-fracn3=frac4n^3-n3.$$
answered 2 days ago
Michael Rozenberg
86.9k1576178
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Your approach is almost correct. Check again your steps. At the end you should have
$$beginalign(k+mathbf2)^3-1
&=6(1+3^2+dots+k^2)+12(1+3+dots+k)+underbrace(8+8+dots+8)_text$(k+1)/2$ times\
&=6(1+3^2+dots+k^2)+12left(frack+12right)^2+8mathbfleft(frack+12right).endalign$$
Hence
$$6(1+3^2+dots+k^2)=(k+mathbf2)^3-1-12left(frack+12right)^2-8mathbfleft(frack+12right)$$
and it follows that
$$sum_j=1^n(2j-1)^2=1+3^2+dots+k^2=frack(k+2)(k+1)6=fracn (4n^2 - 1)3$$
where $n=(k+1)/2$.
edited 2 days ago
Jyrki Lahtonen
104k12161355
answered 2 days ago
Robert Z
83.4k954122
If I was to find the same sum, but instead with even numbers, would I write: 6(1 + 3^2 + ⋯ + k^2) = (k+2)^3 − 2 − 12((k^2)/4 + k/2) − 8((k+1)/2)
– Cameron Choi
2 days ago
@CameronChoi It should be $6(2^2 + dots + k^2) = (k+2)^3 − 2^3 − 12(k^2/4 + k/2) − 8(k/2) $ where $n=k/2$.
– Robert Z
2 days ago
add a comment |Â
up vote
4
down vote
accepted
Your approach is almost correct. Check again your steps. At the end you should have
$$beginalign(k+mathbf2)^3-1
&=6(1+3^2+dots+k^2)+12(1+3+dots+k)+underbrace(8+8+dots+8)_text$(k+1)/2$ times\
&=6(1+3^2+dots+k^2)+12left(frack+12right)^2+8mathbfleft(frack+12right).endalign$$
Hence
$$6(1+3^2+dots+k^2)=(k+mathbf2)^3-1-12left(frack+12right)^2-8mathbfleft(frack+12right)$$
and it follows that
$$sum_j=1^n(2j-1)^2=1+3^2+dots+k^2=frack(k+2)(k+1)6=fracn (4n^2 - 1)3$$
where $n=(k+1)/2$.
edited 2 days ago
Jyrki Lahtonen
104k12161355
answered 2 days ago
Robert Z
83.4k954122
If I was to find the same sum, but instead with even numbers, would I write: 6(1 + 3^2 + ⋯ + k^2) = (k+2)^3 − 2 − 12((k^2)/4 + k/2) − 8((k+1)/2)
– Cameron Choi
2 days ago
@CameronChoi It should be $6(2^2 + dots + k^2) = (k+2)^3 − 2^3 − 12(k^2/4 + k/2) − 8(k/2) $ where $n=k/2$.
– Robert Z
2 days ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Your approach is almost correct. Check again your steps. At the end you should have
$$beginalign(k+mathbf2)^3-1
&=6(1+3^2+dots+k^2)+12(1+3+dots+k)+underbrace(8+8+dots+8)_text$(k+1)/2$ times\
&=6(1+3^2+dots+k^2)+12left(frack+12right)^2+8mathbfleft(frack+12right).endalign$$
Hence
$$6(1+3^2+dots+k^2)=(k+mathbf2)^3-1-12left(frack+12right)^2-8mathbfleft(frack+12right)$$
and it follows that
$$sum_j=1^n(2j-1)^2=1+3^2+dots+k^2=frack(k+2)(k+1)6=fracn (4n^2 - 1)3$$
where $n=(k+1)/2$.
edited 2 days ago
Jyrki Lahtonen
104k12161355
answered 2 days ago
Robert Z
83.4k954122
Your approach is almost correct. Check again your steps. At the end you should have
$$beginalign(k+mathbf2)^3-1
&=6(1+3^2+dots+k^2)+12(1+3+dots+k)+underbrace(8+8+dots+8)_text$(k+1)/2$ times\
&=6(1+3^2+dots+k^2)+12left(frack+12right)^2+8mathbfleft(frack+12right).endalign$$
Hence
$$6(1+3^2+dots+k^2)=(k+mathbf2)^3-1-12left(frack+12right)^2-8mathbfleft(frack+12right)$$
and it follows that
$$sum_j=1^n(2j-1)^2=1+3^2+dots+k^2=frack(k+2)(k+1)6=fracn (4n^2 - 1)3$$
where $n=(k+1)/2$.
edited 2 days ago
Jyrki Lahtonen
104k12161355
answered 2 days ago
Robert Z
83.4k954122
edited 2 days ago
Jyrki Lahtonen
104k12161355
edited 2 days ago
Jyrki Lahtonen
104k12161355
edited 2 days ago
Jyrki Lahtonen
104k12161355
104k12161355
answered 2 days ago
Robert Z
83.4k954122
answered 2 days ago
Robert Z
83.4k954122
answered 2 days ago
Robert Z
83.4k954122
83.4k954122
If I was to find the same sum, but instead with even numbers, would I write: 6(1 + 3^2 + ⋯ + k^2) = (k+2)^3 − 2 − 12((k^2)/4 + k/2) − 8((k+1)/2)
– Cameron Choi
2 days ago
@CameronChoi It should be $6(2^2 + dots + k^2) = (k+2)^3 − 2^3 − 12(k^2/4 + k/2) − 8(k/2) $ where $n=k/2$.
– Robert Z
2 days ago
add a comment |Â
If I was to find the same sum, but instead with even numbers, would I write: 6(1 + 3^2 + ⋯ + k^2) = (k+2)^3 − 2 − 12((k^2)/4 + k/2) − 8((k+1)/2)
– Cameron Choi
2 days ago
@CameronChoi It should be $6(2^2 + dots + k^2) = (k+2)^3 − 2^3 − 12(k^2/4 + k/2) − 8(k/2) $ where $n=k/2$.
– Robert Z
2 days ago
If I was to find the same sum, but instead with even numbers, would I write: 6(1 + 3^2 + ⋯ + k^2) = (k+2)^3 − 2 − 12((k^2)/4 + k/2) − 8((k+1)/2)
– Cameron Choi
2 days ago
If I was to find the same sum, but instead with even numbers, would I write: 6(1 + 3^2 + ⋯ + k^2) = (k+2)^3 − 2 − 12((k^2)/4 + k/2) − 8((k+1)/2)
– Cameron Choi
2 days ago
@CameronChoi It should be $6(2^2 + dots + k^2) = (k+2)^3 − 2^3 − 12(k^2/4 + k/2) − 8(k/2) $ where $n=k/2$.
– Robert Z
2 days ago
@CameronChoi It should be $6(2^2 + dots + k^2) = (k+2)^3 − 2^3 − 12(k^2/4 + k/2) − 8(k/2) $ where $n=k/2$.
– Robert Z
2 days ago
add a comment |Â
up vote
1
down vote
Set $n=2m$
$$(2m+2)^3-(2m)^3=24m^2+24m+8=6(2m+1)^2+2$$
$$impliessum_m=0^n(6(2m+1)^2+2)=sum_m=0^n((2m+2)^3-(2m)^3)=sum_m=0^n(f(m+1)-f(m))$$ where $f(m)=(2m+2)^3$
$$6sum_m=0^n(2m+1)^2+2sum_m=0^n1=f(n+1)-f(0)=?$$
answered 2 days ago
lab bhattacharjee
214k14152263
add a comment |Â
up vote
1
down vote
Set $n=2m$
$$(2m+2)^3-(2m)^3=24m^2+24m+8=6(2m+1)^2+2$$
$$impliessum_m=0^n(6(2m+1)^2+2)=sum_m=0^n((2m+2)^3-(2m)^3)=sum_m=0^n(f(m+1)-f(m))$$ where $f(m)=(2m+2)^3$
$$6sum_m=0^n(2m+1)^2+2sum_m=0^n1=f(n+1)-f(0)=?$$
answered 2 days ago
lab bhattacharjee
214k14152263
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Set $n=2m$
$$(2m+2)^3-(2m)^3=24m^2+24m+8=6(2m+1)^2+2$$
$$impliessum_m=0^n(6(2m+1)^2+2)=sum_m=0^n((2m+2)^3-(2m)^3)=sum_m=0^n(f(m+1)-f(m))$$ where $f(m)=(2m+2)^3$
$$6sum_m=0^n(2m+1)^2+2sum_m=0^n1=f(n+1)-f(0)=?$$
answered 2 days ago
lab bhattacharjee
214k14152263
Set $n=2m$
$$(2m+2)^3-(2m)^3=24m^2+24m+8=6(2m+1)^2+2$$
$$impliessum_m=0^n(6(2m+1)^2+2)=sum_m=0^n((2m+2)^3-(2m)^3)=sum_m=0^n(f(m+1)-f(m))$$ where $f(m)=(2m+2)^3$
$$6sum_m=0^n(2m+1)^2+2sum_m=0^n1=f(n+1)-f(0)=?$$
answered 2 days ago
lab bhattacharjee
214k14152263
answered 2 days ago
lab bhattacharjee
214k14152263
answered 2 days ago
lab bhattacharjee
214k14152263
answered 2 days ago
lab bhattacharjee
214k14152263
214k14152263
add a comment |Â
add a comment |Â
up vote
0
down vote
The telescopic sum helps very well:
$$sum_k=1^n(2k-1)^2=sum_k=1^n(4k^2-4k+1)=sum_k=1^nleft(frac43(k^3-(k-1)^3)-frac13right)=$$
$$=frac43(n^3-0)-fracn3=frac4n^3-n3.$$
answered 2 days ago
Michael Rozenberg
86.9k1576178
add a comment |Â
up vote
0
down vote
The telescopic sum helps very well:
$$sum_k=1^n(2k-1)^2=sum_k=1^n(4k^2-4k+1)=sum_k=1^nleft(frac43(k^3-(k-1)^3)-frac13right)=$$
$$=frac43(n^3-0)-fracn3=frac4n^3-n3.$$
answered 2 days ago
Michael Rozenberg
86.9k1576178
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The telescopic sum helps very well:
$$sum_k=1^n(2k-1)^2=sum_k=1^n(4k^2-4k+1)=sum_k=1^nleft(frac43(k^3-(k-1)^3)-frac13right)=$$
$$=frac43(n^3-0)-fracn3=frac4n^3-n3.$$
answered 2 days ago
Michael Rozenberg
86.9k1576178
The telescopic sum helps very well:
$$sum_k=1^n(2k-1)^2=sum_k=1^n(4k^2-4k+1)=sum_k=1^nleft(frac43(k^3-(k-1)^3)-frac13right)=$$
$$=frac43(n^3-0)-fracn3=frac4n^3-n3.$$
answered 2 days ago
Michael Rozenberg
86.9k1576178
answered 2 days ago
Michael Rozenberg
86.9k1576178
answered 2 days ago
Michael Rozenberg
86.9k1576178
answered 2 days ago
Michael Rozenberg
86.9k1576178
86.9k1576178
add a comment |Â
add a comment |Â
Do you know proof by induction?
– rbird
2 days ago
I only know it very vaguely, sorry.
– Cameron Choi
2 days ago
1
Downvoting all "trusted" users who answer an obvious dupe.
– Jyrki Lahtonen
2 days ago
1
@Jyrki Lahtonen OP asked for a verification of the proof. I corrected his/her errors in bold font.
– Robert Z
2 days ago
@RobertZ You may disagree but I am not convinced by the excuse that because this user made a different error from the previous asker we should keep ten versions of an elementary calculation.
– Jyrki Lahtonen
2 days ago