Having a problem with simplifying a sum given that $(k+1)^3-(k-1)^3=6 k^2+2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Set and its Complement are Measure DenseSteinhaus Theorem in $mathbbR^d$Examples of measures that induce certain inclusions in the Lp spaces.Computing moments$sigma$-algebra generated by a set of subsetsI'm having trouble with a proof by inductionGiven measurable mapping $T$, to show that a random variable is measurable with respect to $sigma(T)$Specific Vitali Covering LemmaShow that a given measure is equal to the Lebesgue measure on Borel subsets on $mathbbR$Understanding limit representation of the exponential function
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Having a problem with simplifying a sum given that $(k+1)^3-(k-1)^3=6 k^2+2$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Set and its Complement are Measure DenseSteinhaus Theorem in $mathbbR^d$Examples of measures that induce certain inclusions in the Lp spaces.Computing moments$sigma$-algebra generated by a set of subsetsI'm having trouble with a proof by inductionGiven measurable mapping $T$, to show that a random variable is measurable with respect to $sigma(T)$Specific Vitali Covering LemmaShow that a given measure is equal to the Lebesgue measure on Borel subsets on $mathbbR$Understanding limit representation of the exponential function
$begingroup$
I was able to prove this first part (given in the title) however I am having trouble continuing with the second part of the question:
Any help or suggestions for how to solve this would be greatly appreciated.
:)
measure-theory proof-explanation
edited Mar 27 at 8:16
ablmf
2,58352452
asked Mar 27 at 4:05
KunosKunos
11
$endgroup$
add a comment |
$begingroup$
I was able to prove this first part (given in the title) however I am having trouble continuing with the second part of the question:
Any help or suggestions for how to solve this would be greatly appreciated.
:)
measure-theory proof-explanation
edited Mar 27 at 8:16
ablmf
2,58352452
asked Mar 27 at 4:05
KunosKunos
11
$endgroup$
1
$begingroup$
Welcome to Math Stack Exchange. Are you familiar with Telescoping Series?
$endgroup$
– J. W. Tanner
Mar 27 at 4:07
1
$begingroup$
Hint: $$sum_k=1^nk^2=sum_k=1^nleft[frac(k+1)^3-(k-1)^3-26right]$$
$endgroup$
– Ángel Mario Gallegos
Mar 27 at 4:10
$begingroup$
or $sum_k=1^n left(6k^2+2right)=sum_k=1^n left((k+1)^3-(k-1)^3right)=(n+1)^3+n^3-1^3-0^3$
$endgroup$
– J. W. Tanner
Mar 27 at 4:18
2
$begingroup$
I think the tag of measure theory could be removed.
$endgroup$
– Mann
Mar 27 at 6:07
add a comment |
$begingroup$
I was able to prove this first part (given in the title) however I am having trouble continuing with the second part of the question:
Any help or suggestions for how to solve this would be greatly appreciated.
:)
measure-theory proof-explanation
edited Mar 27 at 8:16
ablmf
2,58352452
asked Mar 27 at 4:05
KunosKunos
11
$endgroup$
I was able to prove this first part (given in the title) however I am having trouble continuing with the second part of the question:
Any help or suggestions for how to solve this would be greatly appreciated.
:)
measure-theory proof-explanation
measure-theory proof-explanation
edited Mar 27 at 8:16
ablmf
2,58352452
asked Mar 27 at 4:05
KunosKunos
11
edited Mar 27 at 8:16
ablmf
2,58352452
asked Mar 27 at 4:05
KunosKunos
11
edited Mar 27 at 8:16
ablmf
2,58352452
edited Mar 27 at 8:16
ablmf
2,58352452
edited Mar 27 at 8:16
ablmf
2,58352452
2,58352452
asked Mar 27 at 4:05
KunosKunos
11
asked Mar 27 at 4:05
KunosKunos
11
asked Mar 27 at 4:05
KunosKunos
11
11
1
$begingroup$
Welcome to Math Stack Exchange. Are you familiar with Telescoping Series?
$endgroup$
– J. W. Tanner
Mar 27 at 4:07
1
$begingroup$
Hint: $$sum_k=1^nk^2=sum_k=1^nleft[frac(k+1)^3-(k-1)^3-26right]$$
$endgroup$
– Ángel Mario Gallegos
Mar 27 at 4:10
$begingroup$
or $sum_k=1^n left(6k^2+2right)=sum_k=1^n left((k+1)^3-(k-1)^3right)=(n+1)^3+n^3-1^3-0^3$
$endgroup$
– J. W. Tanner
Mar 27 at 4:18
2
$begingroup$
I think the tag of measure theory could be removed.
$endgroup$
– Mann
Mar 27 at 6:07
add a comment |
1
$begingroup$
Welcome to Math Stack Exchange. Are you familiar with Telescoping Series?
$endgroup$
– J. W. Tanner
Mar 27 at 4:07
1
$begingroup$
Hint: $$sum_k=1^nk^2=sum_k=1^nleft[frac(k+1)^3-(k-1)^3-26right]$$
$endgroup$
– Ángel Mario Gallegos
Mar 27 at 4:10
$begingroup$
or $sum_k=1^n left(6k^2+2right)=sum_k=1^n left((k+1)^3-(k-1)^3right)=(n+1)^3+n^3-1^3-0^3$
$endgroup$
– J. W. Tanner
Mar 27 at 4:18
2
$begingroup$
I think the tag of measure theory could be removed.
$endgroup$
– Mann
Mar 27 at 6:07
1
1
$begingroup$
Welcome to Math Stack Exchange. Are you familiar with Telescoping Series?
$endgroup$
– J. W. Tanner
Mar 27 at 4:07
$begingroup$
Welcome to Math Stack Exchange. Are you familiar with Telescoping Series?
$endgroup$
– J. W. Tanner
Mar 27 at 4:07
1
1
$begingroup$
Hint: $$sum_k=1^nk^2=sum_k=1^nleft[frac(k+1)^3-(k-1)^3-26right]$$
$endgroup$
– Ángel Mario Gallegos
Mar 27 at 4:10
$begingroup$
Hint: $$sum_k=1^nk^2=sum_k=1^nleft[frac(k+1)^3-(k-1)^3-26right]$$
$endgroup$
– Ángel Mario Gallegos
Mar 27 at 4:10
$begingroup$
or $sum_k=1^n left(6k^2+2right)=sum_k=1^n left((k+1)^3-(k-1)^3right)=(n+1)^3+n^3-1^3-0^3$
$endgroup$
– J. W. Tanner
Mar 27 at 4:18
$begingroup$
or $sum_k=1^n left(6k^2+2right)=sum_k=1^n left((k+1)^3-(k-1)^3right)=(n+1)^3+n^3-1^3-0^3$
$endgroup$
– J. W. Tanner
Mar 27 at 4:18
2
2
$begingroup$
I think the tag of measure theory could be removed.
$endgroup$
– Mann
Mar 27 at 6:07
$begingroup$
I think the tag of measure theory could be removed.
$endgroup$
– Mann
Mar 27 at 6:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We can use the telescopic sum.
$$sum_k=1^n(6k^2+2)=sum_k=1^n((k+1)^3-(k-1)^3)=$$
$$=sum_k=1^n((k+1)^3-k^3+(k^3-(k-1)^3))=(n+1)^3-1+(n^3-0).$$
Can you end it now?
answered Mar 27 at 4:35
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
Hint
Note that $$sum_1^n 6k^2+2=sum_k=1^n(k+1)^3-(k-1)^3\=sum_k=1^n(k+1)^3-k^3+k^3-(k-1)^3\=sum_k=1^n(k+1)^3-k^3\+sum_k=1^nk^3-(k-1)^3$$Now use the telescopic rule.
answered Mar 27 at 8:16
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
We can use the telescopic sum.
$$sum_k=1^n(6k^2+2)=sum_k=1^n((k+1)^3-(k-1)^3)=$$
$$=sum_k=1^n((k+1)^3-k^3+(k^3-(k-1)^3))=(n+1)^3-1+(n^3-0).$$
Can you end it now?
answered Mar 27 at 4:35
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
We can use the telescopic sum.
$$sum_k=1^n(6k^2+2)=sum_k=1^n((k+1)^3-(k-1)^3)=$$
$$=sum_k=1^n((k+1)^3-k^3+(k^3-(k-1)^3))=(n+1)^3-1+(n^3-0).$$
Can you end it now?
answered Mar 27 at 4:35
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
We can use the telescopic sum.
$$sum_k=1^n(6k^2+2)=sum_k=1^n((k+1)^3-(k-1)^3)=$$
$$=sum_k=1^n((k+1)^3-k^3+(k^3-(k-1)^3))=(n+1)^3-1+(n^3-0).$$
Can you end it now?
answered Mar 27 at 4:35
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
We can use the telescopic sum.
$$sum_k=1^n(6k^2+2)=sum_k=1^n((k+1)^3-(k-1)^3)=$$
$$=sum_k=1^n((k+1)^3-k^3+(k^3-(k-1)^3))=(n+1)^3-1+(n^3-0).$$
Can you end it now?
answered Mar 27 at 4:35
Michael RozenbergMichael Rozenberg
111k1897201
answered Mar 27 at 4:35
Michael RozenbergMichael Rozenberg
111k1897201
answered Mar 27 at 4:35
Michael RozenbergMichael Rozenberg
111k1897201
answered Mar 27 at 4:35
Michael RozenbergMichael Rozenberg
111k1897201
111k1897201
add a comment |
add a comment |
$begingroup$
Hint
Note that $$sum_1^n 6k^2+2=sum_k=1^n(k+1)^3-(k-1)^3\=sum_k=1^n(k+1)^3-k^3+k^3-(k-1)^3\=sum_k=1^n(k+1)^3-k^3\+sum_k=1^nk^3-(k-1)^3$$Now use the telescopic rule.
answered Mar 27 at 8:16
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
Note that $$sum_1^n 6k^2+2=sum_k=1^n(k+1)^3-(k-1)^3\=sum_k=1^n(k+1)^3-k^3+k^3-(k-1)^3\=sum_k=1^n(k+1)^3-k^3\+sum_k=1^nk^3-(k-1)^3$$Now use the telescopic rule.
answered Mar 27 at 8:16
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
Note that $$sum_1^n 6k^2+2=sum_k=1^n(k+1)^3-(k-1)^3\=sum_k=1^n(k+1)^3-k^3+k^3-(k-1)^3\=sum_k=1^n(k+1)^3-k^3\+sum_k=1^nk^3-(k-1)^3$$Now use the telescopic rule.
answered Mar 27 at 8:16
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Hint
Note that $$sum_1^n 6k^2+2=sum_k=1^n(k+1)^3-(k-1)^3\=sum_k=1^n(k+1)^3-k^3+k^3-(k-1)^3\=sum_k=1^n(k+1)^3-k^3\+sum_k=1^nk^3-(k-1)^3$$Now use the telescopic rule.
answered Mar 27 at 8:16
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 27 at 8:16
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 27 at 8:16
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 27 at 8:16
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
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1
$begingroup$
Welcome to Math Stack Exchange. Are you familiar with Telescoping Series?
$endgroup$
– J. W. Tanner
Mar 27 at 4:07
1
$begingroup$
Hint: $$sum_k=1^nk^2=sum_k=1^nleft[frac(k+1)^3-(k-1)^3-26right]$$
$endgroup$
– Ángel Mario Gallegos
Mar 27 at 4:10
$begingroup$
or $sum_k=1^n left(6k^2+2right)=sum_k=1^n left((k+1)^3-(k-1)^3right)=(n+1)^3+n^3-1^3-0^3$
$endgroup$
– J. W. Tanner
Mar 27 at 4:18
2
$begingroup$
I think the tag of measure theory could be removed.
$endgroup$
– Mann
Mar 27 at 6:07