Show $frac1k-1+frac 6k-frac7k+1=frac8k-6k(k^2-1)$, deduce $sumlimits_k = 2^nfrac4k-3k(k^2-1)$ [duplicate]
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This question already has an answer here:
If $k > 1$, then $frac1(k - 1)^2-frac1(k + 1)^2=frac4k(k^2 - 1)^2$, hence simplify $sumlimits_k = 2^nfrac k(k^2 - 1)^2$
2 answers
Prove that if $k > 1$, then
$$ dfrac1k - 1 + dfrac6k - dfrac7k + 1 = dfrac8k - 6k(k^2 - 1)$$
Hence simplify
$$ sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) $$
My Work
$$ dfrac1k - 1 + dfrac6k - dfrac7k + 1 = dfrac8k - 6k(k^2 - 1) = dfrac7k - 6k(k - 1) - dfrac7k + 1$$
$$= dfrac(7k - 6)(k + 1) - 7(k(k - 1))k(k - 1)(k + 1)$$
$$= dfrac8k - 6k(k^2 - 1) = 2 left[ dfrac4k - 3k(k^2 -1) right]$$
$$therefore sum_k = 2^n dfrac4k - 3k(k^2 - 1) = dfrac12 sum_k = 2^n left( dfrac1k -1 + dfrac6k - dfrac7k + 1 right)$$
$$= dfrac12 left( sum_k = 2^n dfrac1k -1 + 6 sum_k = 2^n dfrac1k - 7 sum_k = 2^n dfrac1k + 1 right)$$
$$= dfrac12 left( sum_k = 1^n - 1 dfrac1k - 1 + 6 sum_k = 2^n dfrac1k - 7 sum_k = 3^n + 1 dfrac1k + 1 right)$$
I would appreciate it if people could please take the time to review my work and explain how I should proceed.
sequences-and-series summation
edited Aug 25 at 7:16
Did
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asked Aug 25 at 6:06
The Pointer
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Aug 25 at 6:46
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.
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This question already has an answer here:
If $k > 1$, then $frac1(k - 1)^2-frac1(k + 1)^2=frac4k(k^2 - 1)^2$, hence simplify $sumlimits_k = 2^nfrac k(k^2 - 1)^2$
2 answers
Prove that if $k > 1$, then
$$ dfrac1k - 1 + dfrac6k - dfrac7k + 1 = dfrac8k - 6k(k^2 - 1)$$
Hence simplify
$$ sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) $$
My Work
$$ dfrac1k - 1 + dfrac6k - dfrac7k + 1 = dfrac8k - 6k(k^2 - 1) = dfrac7k - 6k(k - 1) - dfrac7k + 1$$
$$= dfrac(7k - 6)(k + 1) - 7(k(k - 1))k(k - 1)(k + 1)$$
$$= dfrac8k - 6k(k^2 - 1) = 2 left[ dfrac4k - 3k(k^2 -1) right]$$
$$therefore sum_k = 2^n dfrac4k - 3k(k^2 - 1) = dfrac12 sum_k = 2^n left( dfrac1k -1 + dfrac6k - dfrac7k + 1 right)$$
$$= dfrac12 left( sum_k = 2^n dfrac1k -1 + 6 sum_k = 2^n dfrac1k - 7 sum_k = 2^n dfrac1k + 1 right)$$
$$= dfrac12 left( sum_k = 1^n - 1 dfrac1k - 1 + 6 sum_k = 2^n dfrac1k - 7 sum_k = 3^n + 1 dfrac1k + 1 right)$$
I would appreciate it if people could please take the time to review my work and explain how I should proceed.
sequences-and-series summation
edited Aug 25 at 7:16
Did
243k23208443
asked Aug 25 at 6:06
The Pointer
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Aug 25 at 6:46
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.
1
matching the formHow is that a match? You can't just assume that $,1/2 = 1,$ and $,7/2 = 3,$.
– dxiv
Aug 25 at 6:09
@dxiv No idea. I just used pattern matching because I didn't know how to proceed. What would you suggest?
– The Pointer
Aug 25 at 6:10
1
@ThePointer "pattern matching"? What's that? The line after "we get" is completely false! But observe: $$8k-6=2(4k-3).$$
– Lord Shark the Unknown
Aug 25 at 6:14
1
$sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)$
– Anurag A
Aug 25 at 6:14
2
Try writing out the first few terms of each sum explicitly - I would suggest not cancelling common factors when writing out sums like this because that can obscure patterns - keep the denominators as they are.
– Mark Bennet
Aug 25 at 6:45
 |Â
show 4 more comments
up vote
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up vote
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down vote
favorite
This question already has an answer here:
If $k > 1$, then $frac1(k - 1)^2-frac1(k + 1)^2=frac4k(k^2 - 1)^2$, hence simplify $sumlimits_k = 2^nfrac k(k^2 - 1)^2$
2 answers
Prove that if $k > 1$, then
$$ dfrac1k - 1 + dfrac6k - dfrac7k + 1 = dfrac8k - 6k(k^2 - 1)$$
Hence simplify
$$ sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) $$
My Work
$$ dfrac1k - 1 + dfrac6k - dfrac7k + 1 = dfrac8k - 6k(k^2 - 1) = dfrac7k - 6k(k - 1) - dfrac7k + 1$$
$$= dfrac(7k - 6)(k + 1) - 7(k(k - 1))k(k - 1)(k + 1)$$
$$= dfrac8k - 6k(k^2 - 1) = 2 left[ dfrac4k - 3k(k^2 -1) right]$$
$$therefore sum_k = 2^n dfrac4k - 3k(k^2 - 1) = dfrac12 sum_k = 2^n left( dfrac1k -1 + dfrac6k - dfrac7k + 1 right)$$
$$= dfrac12 left( sum_k = 2^n dfrac1k -1 + 6 sum_k = 2^n dfrac1k - 7 sum_k = 2^n dfrac1k + 1 right)$$
$$= dfrac12 left( sum_k = 1^n - 1 dfrac1k - 1 + 6 sum_k = 2^n dfrac1k - 7 sum_k = 3^n + 1 dfrac1k + 1 right)$$
I would appreciate it if people could please take the time to review my work and explain how I should proceed.
sequences-and-series summation
edited Aug 25 at 7:16
Did
243k23208443
asked Aug 25 at 6:06
The Pointer
2,7692830
This question already has an answer here:
If $k > 1$, then $frac1(k - 1)^2-frac1(k + 1)^2=frac4k(k^2 - 1)^2$, hence simplify $sumlimits_k = 2^nfrac k(k^2 - 1)^2$
2 answers
Prove that if $k > 1$, then
$$ dfrac1k - 1 + dfrac6k - dfrac7k + 1 = dfrac8k - 6k(k^2 - 1)$$
Hence simplify
$$ sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) $$
My Work
$$ dfrac1k - 1 + dfrac6k - dfrac7k + 1 = dfrac8k - 6k(k^2 - 1) = dfrac7k - 6k(k - 1) - dfrac7k + 1$$
$$= dfrac(7k - 6)(k + 1) - 7(k(k - 1))k(k - 1)(k + 1)$$
$$= dfrac8k - 6k(k^2 - 1) = 2 left[ dfrac4k - 3k(k^2 -1) right]$$
$$therefore sum_k = 2^n dfrac4k - 3k(k^2 - 1) = dfrac12 sum_k = 2^n left( dfrac1k -1 + dfrac6k - dfrac7k + 1 right)$$
$$= dfrac12 left( sum_k = 2^n dfrac1k -1 + 6 sum_k = 2^n dfrac1k - 7 sum_k = 2^n dfrac1k + 1 right)$$
$$= dfrac12 left( sum_k = 1^n - 1 dfrac1k - 1 + 6 sum_k = 2^n dfrac1k - 7 sum_k = 3^n + 1 dfrac1k + 1 right)$$
I would appreciate it if people could please take the time to review my work and explain how I should proceed.
This question already has an answer here:
If $k > 1$, then $frac1(k - 1)^2-frac1(k + 1)^2=frac4k(k^2 - 1)^2$, hence simplify $sumlimits_k = 2^nfrac k(k^2 - 1)^2$
2 answers
sequences-and-series summation
edited Aug 25 at 7:16
Did
243k23208443
asked Aug 25 at 6:06
The Pointer
2,7692830
edited Aug 25 at 7:16
Did
243k23208443
edited Aug 25 at 7:16
Did
243k23208443
edited Aug 25 at 7:16
Did
243k23208443
243k23208443
asked Aug 25 at 6:06
The Pointer
2,7692830
asked Aug 25 at 6:06
The Pointer
2,7692830
asked Aug 25 at 6:06
The Pointer
2,7692830
2,7692830
marked as duplicate by Did sequences-and-series
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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.
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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.
1
matching the formHow is that a match? You can't just assume that $,1/2 = 1,$ and $,7/2 = 3,$.
– dxiv
Aug 25 at 6:09
@dxiv No idea. I just used pattern matching because I didn't know how to proceed. What would you suggest?
– The Pointer
Aug 25 at 6:10
1
@ThePointer "pattern matching"? What's that? The line after "we get" is completely false! But observe: $$8k-6=2(4k-3).$$
– Lord Shark the Unknown
Aug 25 at 6:14
1
$sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)$
– Anurag A
Aug 25 at 6:14
2
Try writing out the first few terms of each sum explicitly - I would suggest not cancelling common factors when writing out sums like this because that can obscure patterns - keep the denominators as they are.
– Mark Bennet
Aug 25 at 6:45
 |Â
show 4 more comments
1
matching the formHow is that a match? You can't just assume that $,1/2 = 1,$ and $,7/2 = 3,$.
– dxiv
Aug 25 at 6:09
@dxiv No idea. I just used pattern matching because I didn't know how to proceed. What would you suggest?
– The Pointer
Aug 25 at 6:10
1
@ThePointer "pattern matching"? What's that? The line after "we get" is completely false! But observe: $$8k-6=2(4k-3).$$
– Lord Shark the Unknown
Aug 25 at 6:14
1
$sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)$
– Anurag A
Aug 25 at 6:14
2
Try writing out the first few terms of each sum explicitly - I would suggest not cancelling common factors when writing out sums like this because that can obscure patterns - keep the denominators as they are.
– Mark Bennet
Aug 25 at 6:45
1
1
matching the form How is that a match? You can't just assume that $,1/2 = 1,$ and $,7/2 = 3,$.– dxiv
Aug 25 at 6:09
matching the form How is that a match? You can't just assume that $,1/2 = 1,$ and $,7/2 = 3,$.– dxiv
Aug 25 at 6:09
@dxiv No idea. I just used pattern matching because I didn't know how to proceed. What would you suggest?
– The Pointer
Aug 25 at 6:10
@dxiv No idea. I just used pattern matching because I didn't know how to proceed. What would you suggest?
– The Pointer
Aug 25 at 6:10
1
1
@ThePointer "pattern matching"? What's that? The line after "we get" is completely false! But observe: $$8k-6=2(4k-3).$$
– Lord Shark the Unknown
Aug 25 at 6:14
@ThePointer "pattern matching"? What's that? The line after "we get" is completely false! But observe: $$8k-6=2(4k-3).$$
– Lord Shark the Unknown
Aug 25 at 6:14
1
1
$sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)$
– Anurag A
Aug 25 at 6:14
$sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)$
– Anurag A
Aug 25 at 6:14
2
2
Try writing out the first few terms of each sum explicitly - I would suggest not cancelling common factors when writing out sums like this because that can obscure patterns - keep the denominators as they are.
– Mark Bennet
Aug 25 at 6:45
Try writing out the first few terms of each sum explicitly - I would suggest not cancelling common factors when writing out sums like this because that can obscure patterns - keep the denominators as they are.
– Mark Bennet
Aug 25 at 6:45
 |Â
show 4 more comments
1 Answer
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beginalign*
sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) & = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)\
&=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)\
&=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac6k + 1-dfrac1k+1 right)\
&=frac12left[sum_k = 2^n left( dfrac1k - 1 -dfrac1k+1right) +6sum_k = 2^n left( dfrac1k - dfrac1k + 1right)right]\
&=frac12left[left(1+frac12-frac1n-frac1n+1right)+6left(frac12-frac1n+1right)right]
endalign*
In the last step, I am using the idea of a telescopic sum. I hope you can take it from here.
answered Aug 25 at 6:44
Anurag A
22.6k12244
Thank you for the clarification.
– The Pointer
Aug 25 at 9:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
beginalign*
sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) & = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)\
&=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)\
&=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac6k + 1-dfrac1k+1 right)\
&=frac12left[sum_k = 2^n left( dfrac1k - 1 -dfrac1k+1right) +6sum_k = 2^n left( dfrac1k - dfrac1k + 1right)right]\
&=frac12left[left(1+frac12-frac1n-frac1n+1right)+6left(frac12-frac1n+1right)right]
endalign*
In the last step, I am using the idea of a telescopic sum. I hope you can take it from here.
answered Aug 25 at 6:44
Anurag A
22.6k12244
Thank you for the clarification.
– The Pointer
Aug 25 at 9:34
add a comment |Â
up vote
1
down vote
accepted
beginalign*
sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) & = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)\
&=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)\
&=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac6k + 1-dfrac1k+1 right)\
&=frac12left[sum_k = 2^n left( dfrac1k - 1 -dfrac1k+1right) +6sum_k = 2^n left( dfrac1k - dfrac1k + 1right)right]\
&=frac12left[left(1+frac12-frac1n-frac1n+1right)+6left(frac12-frac1n+1right)right]
endalign*
In the last step, I am using the idea of a telescopic sum. I hope you can take it from here.
answered Aug 25 at 6:44
Anurag A
22.6k12244
Thank you for the clarification.
– The Pointer
Aug 25 at 9:34
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
beginalign*
sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) & = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)\
&=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)\
&=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac6k + 1-dfrac1k+1 right)\
&=frac12left[sum_k = 2^n left( dfrac1k - 1 -dfrac1k+1right) +6sum_k = 2^n left( dfrac1k - dfrac1k + 1right)right]\
&=frac12left[left(1+frac12-frac1n-frac1n+1right)+6left(frac12-frac1n+1right)right]
endalign*
In the last step, I am using the idea of a telescopic sum. I hope you can take it from here.
answered Aug 25 at 6:44
Anurag A
22.6k12244
beginalign*
sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) & = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)\
&=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)\
&=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac6k + 1-dfrac1k+1 right)\
&=frac12left[sum_k = 2^n left( dfrac1k - 1 -dfrac1k+1right) +6sum_k = 2^n left( dfrac1k - dfrac1k + 1right)right]\
&=frac12left[left(1+frac12-frac1n-frac1n+1right)+6left(frac12-frac1n+1right)right]
endalign*
In the last step, I am using the idea of a telescopic sum. I hope you can take it from here.
answered Aug 25 at 6:44
Anurag A
22.6k12244
answered Aug 25 at 6:44
Anurag A
22.6k12244
answered Aug 25 at 6:44
Anurag A
22.6k12244
answered Aug 25 at 6:44
Anurag A
22.6k12244
22.6k12244
Thank you for the clarification.
– The Pointer
Aug 25 at 9:34
add a comment |Â
Thank you for the clarification.
– The Pointer
Aug 25 at 9:34
Thank you for the clarification.
– The Pointer
Aug 25 at 9:34
Thank you for the clarification.
– The Pointer
Aug 25 at 9:34
add a comment |Â
1
matching the formHow is that a match? You can't just assume that $,1/2 = 1,$ and $,7/2 = 3,$.– dxiv
Aug 25 at 6:09
@dxiv No idea. I just used pattern matching because I didn't know how to proceed. What would you suggest?
– The Pointer
Aug 25 at 6:10
1
@ThePointer "pattern matching"? What's that? The line after "we get" is completely false! But observe: $$8k-6=2(4k-3).$$
– Lord Shark the Unknown
Aug 25 at 6:14
1
$sumlimits_k = 2^n dfrac4k - 3k(k^2 -1) = frac12sumlimits_k = 2^n dfrac8k - 6k(k^2 -1)=frac12sum_k = 2^n left( dfrac1k - 1 + dfrac6k - dfrac7k + 1 right)$
– Anurag A
Aug 25 at 6:14
2
Try writing out the first few terms of each sum explicitly - I would suggest not cancelling common factors when writing out sums like this because that can obscure patterns - keep the denominators as they are.
– Mark Bennet
Aug 25 at 6:45