Prove that...
up vote
3
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Use Principle of Mathematical Induction to show that, for every integer $nge2$,
$$sum_{{a_{n-1}}=1}^{a_n},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=frac{prodlimits_{i=0}^{n-1},(a_n+i)}{n!},.$$
I have totally no idea to deal with the product. I can understand it when it only uses 3 sigma notation.
summation induction products
edited Nov 22 at 17:51
Batominovski
33.5k33292
asked Nov 19 at 13:05
yuanming luo
233
add a comment |
up vote
3
down vote
favorite
Use Principle of Mathematical Induction to show that, for every integer $nge2$,
$$sum_{{a_{n-1}}=1}^{a_n},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=frac{prodlimits_{i=0}^{n-1},(a_n+i)}{n!},.$$
I have totally no idea to deal with the product. I can understand it when it only uses 3 sigma notation.
summation induction products
edited Nov 22 at 17:51
Batominovski
33.5k33292
asked Nov 19 at 13:05
yuanming luo
233
any hint will be thankful
– yuanming luo
Nov 19 at 13:12
Could anyone check whether it is correct?
– yuanming luo
Nov 19 at 13:14
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
Nov 19 at 13:30
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
Nov 19 at 13:32
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
Nov 19 at 13:41
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Use Principle of Mathematical Induction to show that, for every integer $nge2$,
$$sum_{{a_{n-1}}=1}^{a_n},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=frac{prodlimits_{i=0}^{n-1},(a_n+i)}{n!},.$$
I have totally no idea to deal with the product. I can understand it when it only uses 3 sigma notation.
summation induction products
edited Nov 22 at 17:51
Batominovski
33.5k33292
asked Nov 19 at 13:05
yuanming luo
233
Use Principle of Mathematical Induction to show that, for every integer $nge2$,
$$sum_{{a_{n-1}}=1}^{a_n},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=frac{prodlimits_{i=0}^{n-1},(a_n+i)}{n!},.$$
I have totally no idea to deal with the product. I can understand it when it only uses 3 sigma notation.
summation induction products
summation induction products
edited Nov 22 at 17:51
Batominovski
33.5k33292
asked Nov 19 at 13:05
yuanming luo
233
edited Nov 22 at 17:51
Batominovski
33.5k33292
asked Nov 19 at 13:05
yuanming luo
233
edited Nov 22 at 17:51
Batominovski
33.5k33292
edited Nov 22 at 17:51
Batominovski
33.5k33292
edited Nov 22 at 17:51
Batominovski
33.5k33292
33.5k33292
asked Nov 19 at 13:05
yuanming luo
233
asked Nov 19 at 13:05
yuanming luo
233
asked Nov 19 at 13:05
yuanming luo
233
233
any hint will be thankful
– yuanming luo
Nov 19 at 13:12
Could anyone check whether it is correct?
– yuanming luo
Nov 19 at 13:14
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
Nov 19 at 13:30
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
Nov 19 at 13:32
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
Nov 19 at 13:41
add a comment |
any hint will be thankful
– yuanming luo
Nov 19 at 13:12
Could anyone check whether it is correct?
– yuanming luo
Nov 19 at 13:14
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
Nov 19 at 13:30
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
Nov 19 at 13:32
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
Nov 19 at 13:41
any hint will be thankful
– yuanming luo
Nov 19 at 13:12
any hint will be thankful
– yuanming luo
Nov 19 at 13:12
Could anyone check whether it is correct?
– yuanming luo
Nov 19 at 13:14
Could anyone check whether it is correct?
– yuanming luo
Nov 19 at 13:14
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
Nov 19 at 13:30
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
Nov 19 at 13:30
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
Nov 19 at 13:32
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
Nov 19 at 13:32
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
Nov 19 at 13:41
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
Nov 19 at 13:41
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The correct statement is $$sum_{{a_{n-1}}=1}^{a_n},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=frac{prodlimits_{i=0}^{n-1},(a_n+i)}{n!}text{ for all }ninmathbb{Z}_{>0}text{ and }a_ninmathbb{Z}_{geq 0},.$$ This is equivalent to saying that
$$sum_{{a_{n-1}}=1}^{m},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=binom{m+n-1}{n}text{ for all }minmathbb{Z}_{geq 0}text{ and }ninmathbb{Z}_{>0},.tag{*}$$
To prove by induction, you can justify the Hockey-Stick Identity $$sum_{k=1}^N,binom{k}{r}=binom{N+1}{r+1}text{ for all }N,rinmathbb{Z}_{geq 0},.tag{#}$$
(Therefore, if you want to prove everything from the Hockey-Stick Identity to your own identity, then you need two induction proofs. First, prove (#) by induction on $N$. Then prove (*) by induction on $n$.)
Now that I have given a hint on how to do the job inductively, I present a combinatorial approach. Rewrite (*) as
$$sum_{{a_{n-1}}=1}^{m},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},sum_{a_0=1}^{a_1},1=binom{m+n-1}{n}text{ for all }minmathbb{Z}_{geq 0}text{ and }ninmathbb{Z}_{>0},.$$
The left-hand side of the equation above counts the number of $n$-tuples $left(a_0,a_1,a_2,ldots,a_{n-1}right)$ of integers $a_0,a_1,a_2,ldots,a_{n-1}$ such that $$1=:a_{-1}leq a_0 leq a_1 leq a_2leq ldots leq a_{n-1}leq a_n:=m,.$$
Let $x_j:=a_j-a_{j-1}$ for $j=0,1,2,ldots,n$. Then, $(x_0,x_1,x_2,ldots,x_n)$ is an $(n+1)$-tuple of nonnegative integers such that $$x_0+x_1+x_2+ldots+x_n=m-1,.$$
(We can get $a_0,a_1,a_2,ldots,a_{n-1}$ back by noting that $a_j=1+x_0+x_1+ldots+x_j$ for every $j=0,1,2,ldots,n-1$.
By the stars-and-bars technique, there are precisely
$$binom{(m-1)+(n+1)-1}{(n+1)-1}=binom{m+n-1}{n}$$ solutions $(x_0,x_1,ldots,x_n)$. This proves (*).
answered Nov 22 at 17:30
Batominovski
33.5k33292
if I want to do it for every a belongs to Real Number, I cannot use the identities above.
– yuanming luo
Nov 27 at 10:13
2
How are you going to take the sum if the variables are not integers? This has to be specified.
– Batominovski
Nov 27 at 10:21
2
And how are you going to use induction in nonintegral cases?
– Batominovski
Nov 27 at 10:23
if I replace the a1 by f(a1), it also can do the sum.
– yuanming luo
Nov 27 at 10:46
3
How does the sum work? And which occurrence of $a_1$ do you replace by $f(a_1)$ (there are two occurrences of $a_1$ in your expression)? I think you better ask a separate question, and be clear what you want. At this moment, you are not making any sense.
– Batominovski
Nov 27 at 12:29
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The correct statement is $$sum_{{a_{n-1}}=1}^{a_n},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=frac{prodlimits_{i=0}^{n-1},(a_n+i)}{n!}text{ for all }ninmathbb{Z}_{>0}text{ and }a_ninmathbb{Z}_{geq 0},.$$ This is equivalent to saying that
$$sum_{{a_{n-1}}=1}^{m},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=binom{m+n-1}{n}text{ for all }minmathbb{Z}_{geq 0}text{ and }ninmathbb{Z}_{>0},.tag{*}$$
To prove by induction, you can justify the Hockey-Stick Identity $$sum_{k=1}^N,binom{k}{r}=binom{N+1}{r+1}text{ for all }N,rinmathbb{Z}_{geq 0},.tag{#}$$
(Therefore, if you want to prove everything from the Hockey-Stick Identity to your own identity, then you need two induction proofs. First, prove (#) by induction on $N$. Then prove (*) by induction on $n$.)
Now that I have given a hint on how to do the job inductively, I present a combinatorial approach. Rewrite (*) as
$$sum_{{a_{n-1}}=1}^{m},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},sum_{a_0=1}^{a_1},1=binom{m+n-1}{n}text{ for all }minmathbb{Z}_{geq 0}text{ and }ninmathbb{Z}_{>0},.$$
The left-hand side of the equation above counts the number of $n$-tuples $left(a_0,a_1,a_2,ldots,a_{n-1}right)$ of integers $a_0,a_1,a_2,ldots,a_{n-1}$ such that $$1=:a_{-1}leq a_0 leq a_1 leq a_2leq ldots leq a_{n-1}leq a_n:=m,.$$
Let $x_j:=a_j-a_{j-1}$ for $j=0,1,2,ldots,n$. Then, $(x_0,x_1,x_2,ldots,x_n)$ is an $(n+1)$-tuple of nonnegative integers such that $$x_0+x_1+x_2+ldots+x_n=m-1,.$$
(We can get $a_0,a_1,a_2,ldots,a_{n-1}$ back by noting that $a_j=1+x_0+x_1+ldots+x_j$ for every $j=0,1,2,ldots,n-1$.
By the stars-and-bars technique, there are precisely
$$binom{(m-1)+(n+1)-1}{(n+1)-1}=binom{m+n-1}{n}$$ solutions $(x_0,x_1,ldots,x_n)$. This proves (*).
answered Nov 22 at 17:30
Batominovski
33.5k33292
if I want to do it for every a belongs to Real Number, I cannot use the identities above.
– yuanming luo
Nov 27 at 10:13
2
How are you going to take the sum if the variables are not integers? This has to be specified.
– Batominovski
Nov 27 at 10:21
2
And how are you going to use induction in nonintegral cases?
– Batominovski
Nov 27 at 10:23
if I replace the a1 by f(a1), it also can do the sum.
– yuanming luo
Nov 27 at 10:46
3
How does the sum work? And which occurrence of $a_1$ do you replace by $f(a_1)$ (there are two occurrences of $a_1$ in your expression)? I think you better ask a separate question, and be clear what you want. At this moment, you are not making any sense.
– Batominovski
Nov 27 at 12:29
add a comment |
up vote
2
down vote
accepted
The correct statement is $$sum_{{a_{n-1}}=1}^{a_n},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=frac{prodlimits_{i=0}^{n-1},(a_n+i)}{n!}text{ for all }ninmathbb{Z}_{>0}text{ and }a_ninmathbb{Z}_{geq 0},.$$ This is equivalent to saying that
$$sum_{{a_{n-1}}=1}^{m},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=binom{m+n-1}{n}text{ for all }minmathbb{Z}_{geq 0}text{ and }ninmathbb{Z}_{>0},.tag{*}$$
To prove by induction, you can justify the Hockey-Stick Identity $$sum_{k=1}^N,binom{k}{r}=binom{N+1}{r+1}text{ for all }N,rinmathbb{Z}_{geq 0},.tag{#}$$
(Therefore, if you want to prove everything from the Hockey-Stick Identity to your own identity, then you need two induction proofs. First, prove (#) by induction on $N$. Then prove (*) by induction on $n$.)
Now that I have given a hint on how to do the job inductively, I present a combinatorial approach. Rewrite (*) as
$$sum_{{a_{n-1}}=1}^{m},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},sum_{a_0=1}^{a_1},1=binom{m+n-1}{n}text{ for all }minmathbb{Z}_{geq 0}text{ and }ninmathbb{Z}_{>0},.$$
The left-hand side of the equation above counts the number of $n$-tuples $left(a_0,a_1,a_2,ldots,a_{n-1}right)$ of integers $a_0,a_1,a_2,ldots,a_{n-1}$ such that $$1=:a_{-1}leq a_0 leq a_1 leq a_2leq ldots leq a_{n-1}leq a_n:=m,.$$
Let $x_j:=a_j-a_{j-1}$ for $j=0,1,2,ldots,n$. Then, $(x_0,x_1,x_2,ldots,x_n)$ is an $(n+1)$-tuple of nonnegative integers such that $$x_0+x_1+x_2+ldots+x_n=m-1,.$$
(We can get $a_0,a_1,a_2,ldots,a_{n-1}$ back by noting that $a_j=1+x_0+x_1+ldots+x_j$ for every $j=0,1,2,ldots,n-1$.
By the stars-and-bars technique, there are precisely
$$binom{(m-1)+(n+1)-1}{(n+1)-1}=binom{m+n-1}{n}$$ solutions $(x_0,x_1,ldots,x_n)$. This proves (*).
answered Nov 22 at 17:30
Batominovski
33.5k33292
if I want to do it for every a belongs to Real Number, I cannot use the identities above.
– yuanming luo
Nov 27 at 10:13
2
How are you going to take the sum if the variables are not integers? This has to be specified.
– Batominovski
Nov 27 at 10:21
2
And how are you going to use induction in nonintegral cases?
– Batominovski
Nov 27 at 10:23
if I replace the a1 by f(a1), it also can do the sum.
– yuanming luo
Nov 27 at 10:46
3
How does the sum work? And which occurrence of $a_1$ do you replace by $f(a_1)$ (there are two occurrences of $a_1$ in your expression)? I think you better ask a separate question, and be clear what you want. At this moment, you are not making any sense.
– Batominovski
Nov 27 at 12:29
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The correct statement is $$sum_{{a_{n-1}}=1}^{a_n},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=frac{prodlimits_{i=0}^{n-1},(a_n+i)}{n!}text{ for all }ninmathbb{Z}_{>0}text{ and }a_ninmathbb{Z}_{geq 0},.$$ This is equivalent to saying that
$$sum_{{a_{n-1}}=1}^{m},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=binom{m+n-1}{n}text{ for all }minmathbb{Z}_{geq 0}text{ and }ninmathbb{Z}_{>0},.tag{*}$$
To prove by induction, you can justify the Hockey-Stick Identity $$sum_{k=1}^N,binom{k}{r}=binom{N+1}{r+1}text{ for all }N,rinmathbb{Z}_{geq 0},.tag{#}$$
(Therefore, if you want to prove everything from the Hockey-Stick Identity to your own identity, then you need two induction proofs. First, prove (#) by induction on $N$. Then prove (*) by induction on $n$.)
Now that I have given a hint on how to do the job inductively, I present a combinatorial approach. Rewrite (*) as
$$sum_{{a_{n-1}}=1}^{m},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},sum_{a_0=1}^{a_1},1=binom{m+n-1}{n}text{ for all }minmathbb{Z}_{geq 0}text{ and }ninmathbb{Z}_{>0},.$$
The left-hand side of the equation above counts the number of $n$-tuples $left(a_0,a_1,a_2,ldots,a_{n-1}right)$ of integers $a_0,a_1,a_2,ldots,a_{n-1}$ such that $$1=:a_{-1}leq a_0 leq a_1 leq a_2leq ldots leq a_{n-1}leq a_n:=m,.$$
Let $x_j:=a_j-a_{j-1}$ for $j=0,1,2,ldots,n$. Then, $(x_0,x_1,x_2,ldots,x_n)$ is an $(n+1)$-tuple of nonnegative integers such that $$x_0+x_1+x_2+ldots+x_n=m-1,.$$
(We can get $a_0,a_1,a_2,ldots,a_{n-1}$ back by noting that $a_j=1+x_0+x_1+ldots+x_j$ for every $j=0,1,2,ldots,n-1$.
By the stars-and-bars technique, there are precisely
$$binom{(m-1)+(n+1)-1}{(n+1)-1}=binom{m+n-1}{n}$$ solutions $(x_0,x_1,ldots,x_n)$. This proves (*).
answered Nov 22 at 17:30
Batominovski
33.5k33292
The correct statement is $$sum_{{a_{n-1}}=1}^{a_n},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=frac{prodlimits_{i=0}^{n-1},(a_n+i)}{n!}text{ for all }ninmathbb{Z}_{>0}text{ and }a_ninmathbb{Z}_{geq 0},.$$ This is equivalent to saying that
$$sum_{{a_{n-1}}=1}^{m},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},a_1=binom{m+n-1}{n}text{ for all }minmathbb{Z}_{geq 0}text{ and }ninmathbb{Z}_{>0},.tag{*}$$
To prove by induction, you can justify the Hockey-Stick Identity $$sum_{k=1}^N,binom{k}{r}=binom{N+1}{r+1}text{ for all }N,rinmathbb{Z}_{geq 0},.tag{#}$$
(Therefore, if you want to prove everything from the Hockey-Stick Identity to your own identity, then you need two induction proofs. First, prove (#) by induction on $N$. Then prove (*) by induction on $n$.)
Now that I have given a hint on how to do the job inductively, I present a combinatorial approach. Rewrite (*) as
$$sum_{{a_{n-1}}=1}^{m},sum_{a_{n-2}=1}^{a_{n-1}},ldots,sum_{a_{1}=1}^{a_2},sum_{a_0=1}^{a_1},1=binom{m+n-1}{n}text{ for all }minmathbb{Z}_{geq 0}text{ and }ninmathbb{Z}_{>0},.$$
The left-hand side of the equation above counts the number of $n$-tuples $left(a_0,a_1,a_2,ldots,a_{n-1}right)$ of integers $a_0,a_1,a_2,ldots,a_{n-1}$ such that $$1=:a_{-1}leq a_0 leq a_1 leq a_2leq ldots leq a_{n-1}leq a_n:=m,.$$
Let $x_j:=a_j-a_{j-1}$ for $j=0,1,2,ldots,n$. Then, $(x_0,x_1,x_2,ldots,x_n)$ is an $(n+1)$-tuple of nonnegative integers such that $$x_0+x_1+x_2+ldots+x_n=m-1,.$$
(We can get $a_0,a_1,a_2,ldots,a_{n-1}$ back by noting that $a_j=1+x_0+x_1+ldots+x_j$ for every $j=0,1,2,ldots,n-1$.
By the stars-and-bars technique, there are precisely
$$binom{(m-1)+(n+1)-1}{(n+1)-1}=binom{m+n-1}{n}$$ solutions $(x_0,x_1,ldots,x_n)$. This proves (*).
answered Nov 22 at 17:30
Batominovski
33.5k33292
edited Nov 22 at 17:52
answered Nov 22 at 17:30
Batominovski
33.5k33292
answered Nov 22 at 17:30
Batominovski
33.5k33292
answered Nov 22 at 17:30
Batominovski
33.5k33292
33.5k33292
if I want to do it for every a belongs to Real Number, I cannot use the identities above.
– yuanming luo
Nov 27 at 10:13
2
How are you going to take the sum if the variables are not integers? This has to be specified.
– Batominovski
Nov 27 at 10:21
2
And how are you going to use induction in nonintegral cases?
– Batominovski
Nov 27 at 10:23
if I replace the a1 by f(a1), it also can do the sum.
– yuanming luo
Nov 27 at 10:46
3
How does the sum work? And which occurrence of $a_1$ do you replace by $f(a_1)$ (there are two occurrences of $a_1$ in your expression)? I think you better ask a separate question, and be clear what you want. At this moment, you are not making any sense.
– Batominovski
Nov 27 at 12:29
add a comment |
if I want to do it for every a belongs to Real Number, I cannot use the identities above.
– yuanming luo
Nov 27 at 10:13
2
How are you going to take the sum if the variables are not integers? This has to be specified.
– Batominovski
Nov 27 at 10:21
2
And how are you going to use induction in nonintegral cases?
– Batominovski
Nov 27 at 10:23
if I replace the a1 by f(a1), it also can do the sum.
– yuanming luo
Nov 27 at 10:46
3
How does the sum work? And which occurrence of $a_1$ do you replace by $f(a_1)$ (there are two occurrences of $a_1$ in your expression)? I think you better ask a separate question, and be clear what you want. At this moment, you are not making any sense.
– Batominovski
Nov 27 at 12:29
if I want to do it for every a belongs to Real Number, I cannot use the identities above.
– yuanming luo
Nov 27 at 10:13
if I want to do it for every a belongs to Real Number, I cannot use the identities above.
– yuanming luo
Nov 27 at 10:13
2
2
How are you going to take the sum if the variables are not integers? This has to be specified.
– Batominovski
Nov 27 at 10:21
How are you going to take the sum if the variables are not integers? This has to be specified.
– Batominovski
Nov 27 at 10:21
2
2
And how are you going to use induction in nonintegral cases?
– Batominovski
Nov 27 at 10:23
And how are you going to use induction in nonintegral cases?
– Batominovski
Nov 27 at 10:23
if I replace the a1 by f(a1), it also can do the sum.
– yuanming luo
Nov 27 at 10:46
if I replace the a1 by f(a1), it also can do the sum.
– yuanming luo
Nov 27 at 10:46
3
3
How does the sum work? And which occurrence of $a_1$ do you replace by $f(a_1)$ (there are two occurrences of $a_1$ in your expression)? I think you better ask a separate question, and be clear what you want. At this moment, you are not making any sense.
– Batominovski
Nov 27 at 12:29
How does the sum work? And which occurrence of $a_1$ do you replace by $f(a_1)$ (there are two occurrences of $a_1$ in your expression)? I think you better ask a separate question, and be clear what you want. At this moment, you are not making any sense.
– Batominovski
Nov 27 at 12:29
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any hint will be thankful
– yuanming luo
Nov 19 at 13:12
Could anyone check whether it is correct?
– yuanming luo
Nov 19 at 13:14
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
Nov 19 at 13:30
@swapedoc How to prove it without PMI? Could you give me a hint? That may help.
– yuanming luo
Nov 19 at 13:32
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
Nov 19 at 13:41