Mixing
Use PMI to prove this identity
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Use Principle of Mathematical Induction to show that:
for $nge2$
$$sum_{{a_{n-1}}spacespace=1}^{a_n}sum_{a_{n-2}spacespace=1}^{a_{n-1}}......sum_{a_{1}=1}^{a_2}a_1=frac{prod_{i=0}^{n}(a_n+i)}{n!}$$
I have totally no idea to deal with the product. I can understand it when it only uses 3 sigma notation.
summation products
asked yesterday
yuanming luo
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up vote
0
down vote
favorite
1
Use Principle of Mathematical Induction to show that:
for $nge2$
$$sum_{{a_{n-1}}spacespace=1}^{a_n}sum_{a_{n-2}spacespace=1}^{a_{n-1}}......sum_{a_{1}=1}^{a_2}a_1=frac{prod_{i=0}^{n}(a_n+i)}{n!}$$
I have totally no idea to deal with the product. I can understand it when it only uses 3 sigma notation.
summation products
asked yesterday
yuanming luo
61
New contributor
yuanming luo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
any hint will be thankful
– yuanming luo
yesterday
Could anyone check whether it is correct?
– yuanming luo
yesterday
I think one of the components of PMI lies on whether n=1 is satisfied, your lhs doesnt work on it , so no point in applying .For n>=2 this identity holds true ,I can prove it but not using PMI
– swapedoc
yesterday
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
yesterday
hint: look for faulhaber formula , also try to guess coefficients of x^k in (x+1)(x+2)(x+3)....(x+n) where k can be from 0 to n
– swapedoc
yesterday
add a comment |
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
Use Principle of Mathematical Induction to show that:
for $nge2$
$$sum_{{a_{n-1}}spacespace=1}^{a_n}sum_{a_{n-2}spacespace=1}^{a_{n-1}}......sum_{a_{1}=1}^{a_2}a_1=frac{prod_{i=0}^{n}(a_n+i)}{n!}$$
I have totally no idea to deal with the product. I can understand it when it only uses 3 sigma notation.
summation products
asked yesterday
yuanming luo
61
New contributor
yuanming luo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Use Principle of Mathematical Induction to show that:
for $nge2$
$$sum_{{a_{n-1}}spacespace=1}^{a_n}sum_{a_{n-2}spacespace=1}^{a_{n-1}}......sum_{a_{1}=1}^{a_2}a_1=frac{prod_{i=0}^{n}(a_n+i)}{n!}$$
I have totally no idea to deal with the product. I can understand it when it only uses 3 sigma notation.
summation products
summation products
asked yesterday
yuanming luo
61
New contributor
yuanming luo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
yuanming luo
61
New contributor
yuanming luo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked yesterday
yuanming luo
61
New contributor
yuanming luo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
yuanming luo
61
asked yesterday
yuanming luo
61
61
New contributor
yuanming luo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
yuanming luo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
yuanming luo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
any hint will be thankful
– yuanming luo
yesterday
Could anyone check whether it is correct?
– yuanming luo
yesterday
I think one of the components of PMI lies on whether n=1 is satisfied, your lhs doesnt work on it , so no point in applying .For n>=2 this identity holds true ,I can prove it but not using PMI
– swapedoc
yesterday
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
yesterday
hint: look for faulhaber formula , also try to guess coefficients of x^k in (x+1)(x+2)(x+3)....(x+n) where k can be from 0 to n
– swapedoc
yesterday
add a comment |
any hint will be thankful
– yuanming luo
yesterday
Could anyone check whether it is correct?
– yuanming luo
yesterday
I think one of the components of PMI lies on whether n=1 is satisfied, your lhs doesnt work on it , so no point in applying .For n>=2 this identity holds true ,I can prove it but not using PMI
– swapedoc
yesterday
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
yesterday
hint: look for faulhaber formula , also try to guess coefficients of x^k in (x+1)(x+2)(x+3)....(x+n) where k can be from 0 to n
– swapedoc
yesterday
any hint will be thankful
– yuanming luo
yesterday
any hint will be thankful
– yuanming luo
yesterday
Could anyone check whether it is correct?
– yuanming luo
yesterday
Could anyone check whether it is correct?
– yuanming luo
yesterday
I think one of the components of PMI lies on whether n=1 is satisfied, your lhs doesnt work on it , so no point in applying .For n>=2 this identity holds true ,I can prove it but not using PMI
– swapedoc
yesterday
I think one of the components of PMI lies on whether n=1 is satisfied, your lhs doesnt work on it , so no point in applying .For n>=2 this identity holds true ,I can prove it but not using PMI
– swapedoc
yesterday
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
yesterday
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
yesterday
hint: look for faulhaber formula , also try to guess coefficients of x^k in (x+1)(x+2)(x+3)....(x+n) where k can be from 0 to n
– swapedoc
yesterday
hint: look for faulhaber formula , also try to guess coefficients of x^k in (x+1)(x+2)(x+3)....(x+n) where k can be from 0 to n
– swapedoc
yesterday
add a comment |
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any hint will be thankful
– yuanming luo
yesterday
Could anyone check whether it is correct?
– yuanming luo
yesterday
I think one of the components of PMI lies on whether n=1 is satisfied, your lhs doesnt work on it , so no point in applying .For n>=2 this identity holds true ,I can prove it but not using PMI
– swapedoc
yesterday
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
yesterday
hint: look for faulhaber formula , also try to guess coefficients of x^k in (x+1)(x+2)(x+3)....(x+n) where k can be from 0 to n
– swapedoc
yesterday