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Summation problem involving odd numbers and binomial coefficients. Prove $sum_k=0^mbinomnk(n-2k)(-1)^k=0$
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Recently I came across a finite sum which appears to be zero for all odd numbers. The sum is defined as follows:
$$sum_k=0^m~binomnk(n-2k)(-1)^k$$
where $n=2m+1$.
For the first few $m$ this sum always equals zero. I tried to prove this by induction with the inductive step from $m=k$ to $m=k+1$ but I did not got this far with this approach. So I am asking for a proof of this formula with an explanation.
I have found more of these sums and I would be interested in a more general result. It seems to me like that in general the exponent of the term $n-2k$ is irrelevant as long as it is an odd number. So that
$$sum_k=0^m~binomnk(n-2k)^3(-1)^k$$
$$sum_k=0^m~binomnk(n-2k)^5(-1)^k$$
$$...$$
all equal zero. I guess thats a fact but I have no clue how to derivate this.
I have to add that it is not needed for the number to be odd or even - all these sums should work out for both. So also the following should equal zero.
$$sum_k=0^m~binomnk(n-2k)^2(-1)^k$$
$$sum_k=0^m~binomnk(n-2k)^4(-1)^k$$
$$...$$
summation induction
asked Jul 19 at 14:20
mrtaurho
700219
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up vote
0
down vote
favorite
Recently I came across a finite sum which appears to be zero for all odd numbers. The sum is defined as follows:
$$sum_k=0^m~binomnk(n-2k)(-1)^k$$
where $n=2m+1$.
For the first few $m$ this sum always equals zero. I tried to prove this by induction with the inductive step from $m=k$ to $m=k+1$ but I did not got this far with this approach. So I am asking for a proof of this formula with an explanation.
I have found more of these sums and I would be interested in a more general result. It seems to me like that in general the exponent of the term $n-2k$ is irrelevant as long as it is an odd number. So that
$$sum_k=0^m~binomnk(n-2k)^3(-1)^k$$
$$sum_k=0^m~binomnk(n-2k)^5(-1)^k$$
$$...$$
all equal zero. I guess thats a fact but I have no clue how to derivate this.
I have to add that it is not needed for the number to be odd or even - all these sums should work out for both. So also the following should equal zero.
$$sum_k=0^m~binomnk(n-2k)^2(-1)^k$$
$$sum_k=0^m~binomnk(n-2k)^4(-1)^k$$
$$...$$
summation induction
asked Jul 19 at 14:20
mrtaurho
700219
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Recently I came across a finite sum which appears to be zero for all odd numbers. The sum is defined as follows:
$$sum_k=0^m~binomnk(n-2k)(-1)^k$$
where $n=2m+1$.
For the first few $m$ this sum always equals zero. I tried to prove this by induction with the inductive step from $m=k$ to $m=k+1$ but I did not got this far with this approach. So I am asking for a proof of this formula with an explanation.
I have found more of these sums and I would be interested in a more general result. It seems to me like that in general the exponent of the term $n-2k$ is irrelevant as long as it is an odd number. So that
$$sum_k=0^m~binomnk(n-2k)^3(-1)^k$$
$$sum_k=0^m~binomnk(n-2k)^5(-1)^k$$
$$...$$
all equal zero. I guess thats a fact but I have no clue how to derivate this.
I have to add that it is not needed for the number to be odd or even - all these sums should work out for both. So also the following should equal zero.
$$sum_k=0^m~binomnk(n-2k)^2(-1)^k$$
$$sum_k=0^m~binomnk(n-2k)^4(-1)^k$$
$$...$$
summation induction
asked Jul 19 at 14:20
mrtaurho
700219
Recently I came across a finite sum which appears to be zero for all odd numbers. The sum is defined as follows:
$$sum_k=0^m~binomnk(n-2k)(-1)^k$$
where $n=2m+1$.
For the first few $m$ this sum always equals zero. I tried to prove this by induction with the inductive step from $m=k$ to $m=k+1$ but I did not got this far with this approach. So I am asking for a proof of this formula with an explanation.
I have found more of these sums and I would be interested in a more general result. It seems to me like that in general the exponent of the term $n-2k$ is irrelevant as long as it is an odd number. So that
$$sum_k=0^m~binomnk(n-2k)^3(-1)^k$$
$$sum_k=0^m~binomnk(n-2k)^5(-1)^k$$
$$...$$
all equal zero. I guess thats a fact but I have no clue how to derivate this.
I have to add that it is not needed for the number to be odd or even - all these sums should work out for both. So also the following should equal zero.
$$sum_k=0^m~binomnk(n-2k)^2(-1)^k$$
$$sum_k=0^m~binomnk(n-2k)^4(-1)^k$$
$$...$$
summation induction
asked Jul 19 at 14:20
mrtaurho
700219
edited Jul 19 at 17:59
asked Jul 19 at 14:20
mrtaurho
700219
asked Jul 19 at 14:20
mrtaurho
700219
asked Jul 19 at 14:20
mrtaurho
700219
700219
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
3
down vote
accepted
We have that for $n=2m+1>1$, then
$$beginalign
sum_k=0^m~binomnk(n-2k)(-1)^k&=sum_k=0^m~binomnk(n-k)(-1)^k-sum_k=0^m~binomnkk(-1)^k\
&=sum_k'=n-m^n~binomnn-k'k'(-1)^n-k'-sum_k=0^m~binomnkk(-1)^k\
&=-sum_k'=m+1^n~binomnk'k'(-1)^k'-sum_k=0^m~binomnkk(-1)^k\
&=-sum_k=0^n~binomnkk(-1)^k=nsum_k=1^n~binomn-1k-1(-1)^k-1\
&=n(1+(-1))^n-1=0endalign$$
where at the second step, we rewrite the first sum with respect to the new index $k':=n-k$.
P.S. If $d$ is odd then again by letting $k'=n-k$,
$$sum_k=0^mbinomnk(n-2k)^d(-1)^k=sum_k'=m+1^nbinomnn-k'(n-2(n-k'))^d(-1)^n-k'\=sum_k=m+1^nbinomnk'(n-2k')^d(-1)^k'$$
which implies that
$$sum_k=0^mbinomnk(n-2k)^d(-1)^k=frac12sum_k=0^nbinomnk(n-2k)^d(-1)^k.$$
This sum is NOT always zero. For example, when $n=d$, by Tepper's identity, we have that
$$sum_k=0^nbinomnk(n-2k)^n(-1)^k=2^ncdot n!.$$
answered Jul 19 at 14:28
Robert Z
84.2k954123
I am not quite sure how about the index change from $k=0$ to $k=n-m$ and then further to $k=m+1$. Could you explain this in detail?
– mrtaurho
Jul 19 at 14:48
1
If we replace $k$ with $n-k'$ then the new index $k'=n-k$ goes from $n-0$ to $n-m=m+1$ because $k$ goes from $0$ to $m$.
– Robert Z
Jul 19 at 14:54
Okay, now I realized that this is clear to see. But I am still confused about the substitution of $k$ with $n-k$. I see why do you did it there, but how does $k=0$ become $k=n-m$ and how does $m$ become $n$. My first assumption would be an index shift up by $m$. But how does that effect on the rest of the sum?
– mrtaurho
Jul 19 at 17:17
1
Let the new index be $k'=n-k$. Then for $k=0$, $k'=n$, for $k=1$, $k'=n-1$,..., for $k=m$, $k'=n-m=2m+1-m=m+1$. I edited my answer. Is it better now?
– Robert Z
Jul 19 at 18:16
add a comment |Â
up vote
1
down vote
With the expansion
$$(1-x)^n=sum_k=0^nnchoose k(-x)^k$$
then derivate respect to $x$, and separate the sum into equal number terms:
beginalign
-n(1-x)^n-1
&= sum_k=0^nnchoose k(-k)(-x)^k-1 \
&= sum_k=0^2m+1nchoose k(-k)(-x)^k-1 \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=m+1^2m+1nchoose k(-k)(-x)^k-1 \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=0^mnchoose k+m+1-(k+m+1)(-x)^k+m ~~~ textnow let ~~ kto m-k \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=0^mnchoose 2m+1-k-(2m+1-k)(-x)^2m-k \
&= sum_k=0^mnchoose k(-k)(-1)^k-1x^k-1 + sum_k=0^mnchoose k-(2m+1-k)(-1)^kx^2m-k \
endalign
set $x=1$ then
$$colorblue0= sum_k=0^mnchoose k(k)(-1)^k + sum_k=0^mnchoose k-(2m+1-k)(-1)^k= colorbluesum_k=0^mnchoose k(2k-n)(-1)^k$$
answered Jul 19 at 14:58
Nosrati
19.6k41544
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We have that for $n=2m+1>1$, then
$$beginalign
sum_k=0^m~binomnk(n-2k)(-1)^k&=sum_k=0^m~binomnk(n-k)(-1)^k-sum_k=0^m~binomnkk(-1)^k\
&=sum_k'=n-m^n~binomnn-k'k'(-1)^n-k'-sum_k=0^m~binomnkk(-1)^k\
&=-sum_k'=m+1^n~binomnk'k'(-1)^k'-sum_k=0^m~binomnkk(-1)^k\
&=-sum_k=0^n~binomnkk(-1)^k=nsum_k=1^n~binomn-1k-1(-1)^k-1\
&=n(1+(-1))^n-1=0endalign$$
where at the second step, we rewrite the first sum with respect to the new index $k':=n-k$.
P.S. If $d$ is odd then again by letting $k'=n-k$,
$$sum_k=0^mbinomnk(n-2k)^d(-1)^k=sum_k'=m+1^nbinomnn-k'(n-2(n-k'))^d(-1)^n-k'\=sum_k=m+1^nbinomnk'(n-2k')^d(-1)^k'$$
which implies that
$$sum_k=0^mbinomnk(n-2k)^d(-1)^k=frac12sum_k=0^nbinomnk(n-2k)^d(-1)^k.$$
This sum is NOT always zero. For example, when $n=d$, by Tepper's identity, we have that
$$sum_k=0^nbinomnk(n-2k)^n(-1)^k=2^ncdot n!.$$
answered Jul 19 at 14:28
Robert Z
84.2k954123
I am not quite sure how about the index change from $k=0$ to $k=n-m$ and then further to $k=m+1$. Could you explain this in detail?
– mrtaurho
Jul 19 at 14:48
1
If we replace $k$ with $n-k'$ then the new index $k'=n-k$ goes from $n-0$ to $n-m=m+1$ because $k$ goes from $0$ to $m$.
– Robert Z
Jul 19 at 14:54
Okay, now I realized that this is clear to see. But I am still confused about the substitution of $k$ with $n-k$. I see why do you did it there, but how does $k=0$ become $k=n-m$ and how does $m$ become $n$. My first assumption would be an index shift up by $m$. But how does that effect on the rest of the sum?
– mrtaurho
Jul 19 at 17:17
1
Let the new index be $k'=n-k$. Then for $k=0$, $k'=n$, for $k=1$, $k'=n-1$,..., for $k=m$, $k'=n-m=2m+1-m=m+1$. I edited my answer. Is it better now?
– Robert Z
Jul 19 at 18:16
add a comment |Â
up vote
3
down vote
accepted
We have that for $n=2m+1>1$, then
$$beginalign
sum_k=0^m~binomnk(n-2k)(-1)^k&=sum_k=0^m~binomnk(n-k)(-1)^k-sum_k=0^m~binomnkk(-1)^k\
&=sum_k'=n-m^n~binomnn-k'k'(-1)^n-k'-sum_k=0^m~binomnkk(-1)^k\
&=-sum_k'=m+1^n~binomnk'k'(-1)^k'-sum_k=0^m~binomnkk(-1)^k\
&=-sum_k=0^n~binomnkk(-1)^k=nsum_k=1^n~binomn-1k-1(-1)^k-1\
&=n(1+(-1))^n-1=0endalign$$
where at the second step, we rewrite the first sum with respect to the new index $k':=n-k$.
P.S. If $d$ is odd then again by letting $k'=n-k$,
$$sum_k=0^mbinomnk(n-2k)^d(-1)^k=sum_k'=m+1^nbinomnn-k'(n-2(n-k'))^d(-1)^n-k'\=sum_k=m+1^nbinomnk'(n-2k')^d(-1)^k'$$
which implies that
$$sum_k=0^mbinomnk(n-2k)^d(-1)^k=frac12sum_k=0^nbinomnk(n-2k)^d(-1)^k.$$
This sum is NOT always zero. For example, when $n=d$, by Tepper's identity, we have that
$$sum_k=0^nbinomnk(n-2k)^n(-1)^k=2^ncdot n!.$$
answered Jul 19 at 14:28
Robert Z
84.2k954123
I am not quite sure how about the index change from $k=0$ to $k=n-m$ and then further to $k=m+1$. Could you explain this in detail?
– mrtaurho
Jul 19 at 14:48
1
If we replace $k$ with $n-k'$ then the new index $k'=n-k$ goes from $n-0$ to $n-m=m+1$ because $k$ goes from $0$ to $m$.
– Robert Z
Jul 19 at 14:54
Okay, now I realized that this is clear to see. But I am still confused about the substitution of $k$ with $n-k$. I see why do you did it there, but how does $k=0$ become $k=n-m$ and how does $m$ become $n$. My first assumption would be an index shift up by $m$. But how does that effect on the rest of the sum?
– mrtaurho
Jul 19 at 17:17
1
Let the new index be $k'=n-k$. Then for $k=0$, $k'=n$, for $k=1$, $k'=n-1$,..., for $k=m$, $k'=n-m=2m+1-m=m+1$. I edited my answer. Is it better now?
– Robert Z
Jul 19 at 18:16
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We have that for $n=2m+1>1$, then
$$beginalign
sum_k=0^m~binomnk(n-2k)(-1)^k&=sum_k=0^m~binomnk(n-k)(-1)^k-sum_k=0^m~binomnkk(-1)^k\
&=sum_k'=n-m^n~binomnn-k'k'(-1)^n-k'-sum_k=0^m~binomnkk(-1)^k\
&=-sum_k'=m+1^n~binomnk'k'(-1)^k'-sum_k=0^m~binomnkk(-1)^k\
&=-sum_k=0^n~binomnkk(-1)^k=nsum_k=1^n~binomn-1k-1(-1)^k-1\
&=n(1+(-1))^n-1=0endalign$$
where at the second step, we rewrite the first sum with respect to the new index $k':=n-k$.
P.S. If $d$ is odd then again by letting $k'=n-k$,
$$sum_k=0^mbinomnk(n-2k)^d(-1)^k=sum_k'=m+1^nbinomnn-k'(n-2(n-k'))^d(-1)^n-k'\=sum_k=m+1^nbinomnk'(n-2k')^d(-1)^k'$$
which implies that
$$sum_k=0^mbinomnk(n-2k)^d(-1)^k=frac12sum_k=0^nbinomnk(n-2k)^d(-1)^k.$$
This sum is NOT always zero. For example, when $n=d$, by Tepper's identity, we have that
$$sum_k=0^nbinomnk(n-2k)^n(-1)^k=2^ncdot n!.$$
answered Jul 19 at 14:28
Robert Z
84.2k954123
We have that for $n=2m+1>1$, then
$$beginalign
sum_k=0^m~binomnk(n-2k)(-1)^k&=sum_k=0^m~binomnk(n-k)(-1)^k-sum_k=0^m~binomnkk(-1)^k\
&=sum_k'=n-m^n~binomnn-k'k'(-1)^n-k'-sum_k=0^m~binomnkk(-1)^k\
&=-sum_k'=m+1^n~binomnk'k'(-1)^k'-sum_k=0^m~binomnkk(-1)^k\
&=-sum_k=0^n~binomnkk(-1)^k=nsum_k=1^n~binomn-1k-1(-1)^k-1\
&=n(1+(-1))^n-1=0endalign$$
where at the second step, we rewrite the first sum with respect to the new index $k':=n-k$.
P.S. If $d$ is odd then again by letting $k'=n-k$,
$$sum_k=0^mbinomnk(n-2k)^d(-1)^k=sum_k'=m+1^nbinomnn-k'(n-2(n-k'))^d(-1)^n-k'\=sum_k=m+1^nbinomnk'(n-2k')^d(-1)^k'$$
which implies that
$$sum_k=0^mbinomnk(n-2k)^d(-1)^k=frac12sum_k=0^nbinomnk(n-2k)^d(-1)^k.$$
This sum is NOT always zero. For example, when $n=d$, by Tepper's identity, we have that
$$sum_k=0^nbinomnk(n-2k)^n(-1)^k=2^ncdot n!.$$
answered Jul 19 at 14:28
Robert Z
84.2k954123
edited Jul 19 at 18:57
answered Jul 19 at 14:28
Robert Z
84.2k954123
answered Jul 19 at 14:28
Robert Z
84.2k954123
answered Jul 19 at 14:28
Robert Z
84.2k954123
84.2k954123
I am not quite sure how about the index change from $k=0$ to $k=n-m$ and then further to $k=m+1$. Could you explain this in detail?
– mrtaurho
Jul 19 at 14:48
1
If we replace $k$ with $n-k'$ then the new index $k'=n-k$ goes from $n-0$ to $n-m=m+1$ because $k$ goes from $0$ to $m$.
– Robert Z
Jul 19 at 14:54
Okay, now I realized that this is clear to see. But I am still confused about the substitution of $k$ with $n-k$. I see why do you did it there, but how does $k=0$ become $k=n-m$ and how does $m$ become $n$. My first assumption would be an index shift up by $m$. But how does that effect on the rest of the sum?
– mrtaurho
Jul 19 at 17:17
1
Let the new index be $k'=n-k$. Then for $k=0$, $k'=n$, for $k=1$, $k'=n-1$,..., for $k=m$, $k'=n-m=2m+1-m=m+1$. I edited my answer. Is it better now?
– Robert Z
Jul 19 at 18:16
add a comment |Â
I am not quite sure how about the index change from $k=0$ to $k=n-m$ and then further to $k=m+1$. Could you explain this in detail?
– mrtaurho
Jul 19 at 14:48
1
If we replace $k$ with $n-k'$ then the new index $k'=n-k$ goes from $n-0$ to $n-m=m+1$ because $k$ goes from $0$ to $m$.
– Robert Z
Jul 19 at 14:54
Okay, now I realized that this is clear to see. But I am still confused about the substitution of $k$ with $n-k$. I see why do you did it there, but how does $k=0$ become $k=n-m$ and how does $m$ become $n$. My first assumption would be an index shift up by $m$. But how does that effect on the rest of the sum?
– mrtaurho
Jul 19 at 17:17
1
Let the new index be $k'=n-k$. Then for $k=0$, $k'=n$, for $k=1$, $k'=n-1$,..., for $k=m$, $k'=n-m=2m+1-m=m+1$. I edited my answer. Is it better now?
– Robert Z
Jul 19 at 18:16
I am not quite sure how about the index change from $k=0$ to $k=n-m$ and then further to $k=m+1$. Could you explain this in detail?
– mrtaurho
Jul 19 at 14:48
I am not quite sure how about the index change from $k=0$ to $k=n-m$ and then further to $k=m+1$. Could you explain this in detail?
– mrtaurho
Jul 19 at 14:48
1
1
If we replace $k$ with $n-k'$ then the new index $k'=n-k$ goes from $n-0$ to $n-m=m+1$ because $k$ goes from $0$ to $m$.
– Robert Z
Jul 19 at 14:54
If we replace $k$ with $n-k'$ then the new index $k'=n-k$ goes from $n-0$ to $n-m=m+1$ because $k$ goes from $0$ to $m$.
– Robert Z
Jul 19 at 14:54
Okay, now I realized that this is clear to see. But I am still confused about the substitution of $k$ with $n-k$. I see why do you did it there, but how does $k=0$ become $k=n-m$ and how does $m$ become $n$. My first assumption would be an index shift up by $m$. But how does that effect on the rest of the sum?
– mrtaurho
Jul 19 at 17:17
Okay, now I realized that this is clear to see. But I am still confused about the substitution of $k$ with $n-k$. I see why do you did it there, but how does $k=0$ become $k=n-m$ and how does $m$ become $n$. My first assumption would be an index shift up by $m$. But how does that effect on the rest of the sum?
– mrtaurho
Jul 19 at 17:17
1
1
Let the new index be $k'=n-k$. Then for $k=0$, $k'=n$, for $k=1$, $k'=n-1$,..., for $k=m$, $k'=n-m=2m+1-m=m+1$. I edited my answer. Is it better now?
– Robert Z
Jul 19 at 18:16
Let the new index be $k'=n-k$. Then for $k=0$, $k'=n$, for $k=1$, $k'=n-1$,..., for $k=m$, $k'=n-m=2m+1-m=m+1$. I edited my answer. Is it better now?
– Robert Z
Jul 19 at 18:16
add a comment |Â
up vote
1
down vote
With the expansion
$$(1-x)^n=sum_k=0^nnchoose k(-x)^k$$
then derivate respect to $x$, and separate the sum into equal number terms:
beginalign
-n(1-x)^n-1
&= sum_k=0^nnchoose k(-k)(-x)^k-1 \
&= sum_k=0^2m+1nchoose k(-k)(-x)^k-1 \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=m+1^2m+1nchoose k(-k)(-x)^k-1 \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=0^mnchoose k+m+1-(k+m+1)(-x)^k+m ~~~ textnow let ~~ kto m-k \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=0^mnchoose 2m+1-k-(2m+1-k)(-x)^2m-k \
&= sum_k=0^mnchoose k(-k)(-1)^k-1x^k-1 + sum_k=0^mnchoose k-(2m+1-k)(-1)^kx^2m-k \
endalign
set $x=1$ then
$$colorblue0= sum_k=0^mnchoose k(k)(-1)^k + sum_k=0^mnchoose k-(2m+1-k)(-1)^k= colorbluesum_k=0^mnchoose k(2k-n)(-1)^k$$
answered Jul 19 at 14:58
Nosrati
19.6k41544
add a comment |Â
up vote
1
down vote
With the expansion
$$(1-x)^n=sum_k=0^nnchoose k(-x)^k$$
then derivate respect to $x$, and separate the sum into equal number terms:
beginalign
-n(1-x)^n-1
&= sum_k=0^nnchoose k(-k)(-x)^k-1 \
&= sum_k=0^2m+1nchoose k(-k)(-x)^k-1 \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=m+1^2m+1nchoose k(-k)(-x)^k-1 \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=0^mnchoose k+m+1-(k+m+1)(-x)^k+m ~~~ textnow let ~~ kto m-k \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=0^mnchoose 2m+1-k-(2m+1-k)(-x)^2m-k \
&= sum_k=0^mnchoose k(-k)(-1)^k-1x^k-1 + sum_k=0^mnchoose k-(2m+1-k)(-1)^kx^2m-k \
endalign
set $x=1$ then
$$colorblue0= sum_k=0^mnchoose k(k)(-1)^k + sum_k=0^mnchoose k-(2m+1-k)(-1)^k= colorbluesum_k=0^mnchoose k(2k-n)(-1)^k$$
answered Jul 19 at 14:58
Nosrati
19.6k41544
add a comment |Â
up vote
1
down vote
up vote
1
down vote
With the expansion
$$(1-x)^n=sum_k=0^nnchoose k(-x)^k$$
then derivate respect to $x$, and separate the sum into equal number terms:
beginalign
-n(1-x)^n-1
&= sum_k=0^nnchoose k(-k)(-x)^k-1 \
&= sum_k=0^2m+1nchoose k(-k)(-x)^k-1 \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=m+1^2m+1nchoose k(-k)(-x)^k-1 \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=0^mnchoose k+m+1-(k+m+1)(-x)^k+m ~~~ textnow let ~~ kto m-k \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=0^mnchoose 2m+1-k-(2m+1-k)(-x)^2m-k \
&= sum_k=0^mnchoose k(-k)(-1)^k-1x^k-1 + sum_k=0^mnchoose k-(2m+1-k)(-1)^kx^2m-k \
endalign
set $x=1$ then
$$colorblue0= sum_k=0^mnchoose k(k)(-1)^k + sum_k=0^mnchoose k-(2m+1-k)(-1)^k= colorbluesum_k=0^mnchoose k(2k-n)(-1)^k$$
answered Jul 19 at 14:58
Nosrati
19.6k41544
With the expansion
$$(1-x)^n=sum_k=0^nnchoose k(-x)^k$$
then derivate respect to $x$, and separate the sum into equal number terms:
beginalign
-n(1-x)^n-1
&= sum_k=0^nnchoose k(-k)(-x)^k-1 \
&= sum_k=0^2m+1nchoose k(-k)(-x)^k-1 \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=m+1^2m+1nchoose k(-k)(-x)^k-1 \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=0^mnchoose k+m+1-(k+m+1)(-x)^k+m ~~~ textnow let ~~ kto m-k \
&= sum_k=0^mnchoose k(-k)(-x)^k-1 + sum_k=0^mnchoose 2m+1-k-(2m+1-k)(-x)^2m-k \
&= sum_k=0^mnchoose k(-k)(-1)^k-1x^k-1 + sum_k=0^mnchoose k-(2m+1-k)(-1)^kx^2m-k \
endalign
set $x=1$ then
$$colorblue0= sum_k=0^mnchoose k(k)(-1)^k + sum_k=0^mnchoose k-(2m+1-k)(-1)^k= colorbluesum_k=0^mnchoose k(2k-n)(-1)^k$$
answered Jul 19 at 14:58
Nosrati
19.6k41544
answered Jul 19 at 14:58
Nosrati
19.6k41544
answered Jul 19 at 14:58
Nosrati
19.6k41544
answered Jul 19 at 14:58
Nosrati
19.6k41544
19.6k41544
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