A proving question based on binomial theorem [on hold]
up vote
-2
down vote
favorite
$$C_0-C1(a-1)(b-1)(c-1)_+C_2(a-2)(b-2)(c-2)+.... (-1)^nC_n(a-n)(b-n)(c-n) $$=0 I tried to solve this problem by using multinomial theorem but was not able to proceed further please help me out.
binomial-coefficients binomial-theorem binomial-distribution multinomial-theorem
asked Dec 2 at 3:39
priyanka kumari
1177
put on hold as off-topic by user302797, Paul Frost, Cesareo, Yanko, amWhy Dec 2 at 15:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, Paul Frost, Cesareo, Yanko, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-2
down vote
favorite
$$C_0-C1(a-1)(b-1)(c-1)_+C_2(a-2)(b-2)(c-2)+.... (-1)^nC_n(a-n)(b-n)(c-n) $$=0 I tried to solve this problem by using multinomial theorem but was not able to proceed further please help me out.
binomial-coefficients binomial-theorem binomial-distribution multinomial-theorem
asked Dec 2 at 3:39
priyanka kumari
1177
put on hold as off-topic by user302797, Paul Frost, Cesareo, Yanko, amWhy Dec 2 at 15:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, Paul Frost, Cesareo, Yanko, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
What are the $C_k$ ? What are $a,b,c$?
– darij grinberg
Dec 2 at 3:51
$C_k$ is the binomial coefficient $binom nk$
– Shubham Johri
Dec 2 at 5:23
@priyanka kumari Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4,a=b=c=1/2$, and it seems it should be $abcC_0$.
– Shubham Johri
Dec 2 at 6:56
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
$$C_0-C1(a-1)(b-1)(c-1)_+C_2(a-2)(b-2)(c-2)+.... (-1)^nC_n(a-n)(b-n)(c-n) $$=0 I tried to solve this problem by using multinomial theorem but was not able to proceed further please help me out.
binomial-coefficients binomial-theorem binomial-distribution multinomial-theorem
asked Dec 2 at 3:39
priyanka kumari
1177
$$C_0-C1(a-1)(b-1)(c-1)_+C_2(a-2)(b-2)(c-2)+.... (-1)^nC_n(a-n)(b-n)(c-n) $$=0 I tried to solve this problem by using multinomial theorem but was not able to proceed further please help me out.
binomial-coefficients binomial-theorem binomial-distribution multinomial-theorem
binomial-coefficients binomial-theorem binomial-distribution multinomial-theorem
asked Dec 2 at 3:39
priyanka kumari
1177
asked Dec 2 at 3:39
priyanka kumari
1177
asked Dec 2 at 3:39
priyanka kumari
1177
asked Dec 2 at 3:39
priyanka kumari
1177
asked Dec 2 at 3:39
priyanka kumari
1177
1177
put on hold as off-topic by user302797, Paul Frost, Cesareo, Yanko, amWhy Dec 2 at 15:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, Paul Frost, Cesareo, Yanko, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by user302797, Paul Frost, Cesareo, Yanko, amWhy Dec 2 at 15:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, Paul Frost, Cesareo, Yanko, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
What are the $C_k$ ? What are $a,b,c$?
– darij grinberg
Dec 2 at 3:51
$C_k$ is the binomial coefficient $binom nk$
– Shubham Johri
Dec 2 at 5:23
@priyanka kumari Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4,a=b=c=1/2$, and it seems it should be $abcC_0$.
– Shubham Johri
Dec 2 at 6:56
add a comment |
What are the $C_k$ ? What are $a,b,c$?
– darij grinberg
Dec 2 at 3:51
$C_k$ is the binomial coefficient $binom nk$
– Shubham Johri
Dec 2 at 5:23
@priyanka kumari Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4,a=b=c=1/2$, and it seems it should be $abcC_0$.
– Shubham Johri
Dec 2 at 6:56
What are the $C_k$ ? What are $a,b,c$?
– darij grinberg
Dec 2 at 3:51
What are the $C_k$ ? What are $a,b,c$?
– darij grinberg
Dec 2 at 3:51
$C_k$ is the binomial coefficient $binom nk$
– Shubham Johri
Dec 2 at 5:23
$C_k$ is the binomial coefficient $binom nk$
– Shubham Johri
Dec 2 at 5:23
@priyanka kumari Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4,a=b=c=1/2$, and it seems it should be $abcC_0$.
– Shubham Johri
Dec 2 at 6:56
@priyanka kumari Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4,a=b=c=1/2$, and it seems it should be $abcC_0$.
– Shubham Johri
Dec 2 at 6:56
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
From
begin{align*}
(1-x)^n = sum (-1)^i binom{n}{i} x^i
end{align*}
differentiating, we obtain
begin{align*}
n(1-x)^{n-1} = sum (-1)^ii binom{n}{i} x^{i-1}
end{align*}
Multiply by $x$ and differentiate:
begin{align*}
n(1-x)^{n-1}x &= sum (-1)^ii binom{n}{i} x^{i}\
n(n-1)(1-x)^{n-2}x + n(1-x)^{n-1} = sum (-1)^i i^2 binom{n}{i} x^{i-1}
end{align*}
Use the same technique for obtaining an expression for $sum (-1)^i i^3 x^i binom{n}{i}$.
The given expression can be expanded as
begin{align*}
sum (-1)^i binom{n}{i} [abc - i(a+b+c) + i^2(ab+bc+ca) -i^3]
end{align*}
Put $x=1$ and see that all the individual expressions evaluate to 0.
answered Dec 2 at 4:23
Muralidharan
38516
Not able to understand please tell some other method.
– priyanka kumari
Dec 2 at 5:31
add a comment |
up vote
0
down vote
$(x-1)^n=C_0x^n-C_1x^{n-1}+C_2x^{n-2}...+(-1)^nC_n\$
Multiply by $x^{a-n}$,
$implies (x-1)^nx^{a-n}=C_0x^a-C_1x^{a-1}+C_2x^{a-2}...+(-1)^nC_nx^{a-n}\$
Differentiate with respect to $x$,
$n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}=aC_0x^{a-1}-(a-1)C_1x^{a-2}+(a-2)C_2x^{a-3}...+(-1)^n(a-n)C_nx^{a-n-1}\$
Multiply with $x^{b-a+1}$,
$implies x^{b-a+1}[n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}]=n(x-1)^{n-1}x^{b-n+1}+(a-n)(x-1)^nx^{b-n}=aC_0x^b-(a-1)C_1x^{b-1}+(a-2)C_2x^{b-2}...+(-1)^n(a-n)C_nx^{b-n}\$
Differentiate with respect to $x$,
$n(n-1)(x-1)^{n-2}x^{b-n+1}+n(b-n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}+n(a-n)(x-1)^{n-1}x^{b-n}=abC_0x^{b-1}-(a-1)(b-1)C_1x^{b-2}+(a-2)(b-2)C_2x^{b-3}...+(-1)^n(a-n)(b-n)C_nx^{b-n-1}\$
Multiply with $x^{c-b+1}$,
$x^{c-b+1}[n(n-1)(x-1)^{n-2}x^{b-n+1}+n(a+b-2n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}]=n(n-1)(x-1)^{n-2}x^{c-n+2}+n(a+b-2n+1)(x-1)^{n-1}x^{c-n+1}+(a-n)(b-n)(x-1)^nx^{c-n}=abC_0x^c-(a-1)(b-1)C_1x^{c-1}+(a-2)(b-2)C_2x^{c-2}...+(-1)^n(a-n)(b-n)C_nx^{c-n}\$
Differentiate with respect to $x$,
$n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(c-n+2)(x-1)^{n-2}x^{c-n+1}+n(n-1)(a+b-2n+1)(x-1)^{n-2}x^{c-n+1}+n(c-n+1)(a+b-2n+1)(x-1)^{n-1}x^{c-n}+n(a-n)(b-n)(x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(a+b+c-3n+3)(x-1)^{n-2}x^{c-n+1}+n[(c-n+1)(a+b-2n+1)+(a-n)(b-n)](x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=abcC_0x^{c-1}-(a-1)(b-1)(c-1)C_1x^{c-2}+(a-2)(b-2)(c-2)C_2x^{c-3}...+(-1)^n(a-n)(b-n)(c-n)C_nx^{c-n-1}\$
Set $n>3, x=1$,
$0=abcC_0-(a-1)(b-1)(c-1)C_1+(a-2)(b-2)(c-2)C_2...+(-1)^n(a-n)(b-n)(c-n)C_n\$
Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4, a=b=c=1/2$, and it seems it should be $abcC_0$.
answered Dec 2 at 6:46
Shubham Johri
1,08339
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
From
begin{align*}
(1-x)^n = sum (-1)^i binom{n}{i} x^i
end{align*}
differentiating, we obtain
begin{align*}
n(1-x)^{n-1} = sum (-1)^ii binom{n}{i} x^{i-1}
end{align*}
Multiply by $x$ and differentiate:
begin{align*}
n(1-x)^{n-1}x &= sum (-1)^ii binom{n}{i} x^{i}\
n(n-1)(1-x)^{n-2}x + n(1-x)^{n-1} = sum (-1)^i i^2 binom{n}{i} x^{i-1}
end{align*}
Use the same technique for obtaining an expression for $sum (-1)^i i^3 x^i binom{n}{i}$.
The given expression can be expanded as
begin{align*}
sum (-1)^i binom{n}{i} [abc - i(a+b+c) + i^2(ab+bc+ca) -i^3]
end{align*}
Put $x=1$ and see that all the individual expressions evaluate to 0.
answered Dec 2 at 4:23
Muralidharan
38516
Not able to understand please tell some other method.
– priyanka kumari
Dec 2 at 5:31
add a comment |
up vote
0
down vote
accepted
From
begin{align*}
(1-x)^n = sum (-1)^i binom{n}{i} x^i
end{align*}
differentiating, we obtain
begin{align*}
n(1-x)^{n-1} = sum (-1)^ii binom{n}{i} x^{i-1}
end{align*}
Multiply by $x$ and differentiate:
begin{align*}
n(1-x)^{n-1}x &= sum (-1)^ii binom{n}{i} x^{i}\
n(n-1)(1-x)^{n-2}x + n(1-x)^{n-1} = sum (-1)^i i^2 binom{n}{i} x^{i-1}
end{align*}
Use the same technique for obtaining an expression for $sum (-1)^i i^3 x^i binom{n}{i}$.
The given expression can be expanded as
begin{align*}
sum (-1)^i binom{n}{i} [abc - i(a+b+c) + i^2(ab+bc+ca) -i^3]
end{align*}
Put $x=1$ and see that all the individual expressions evaluate to 0.
answered Dec 2 at 4:23
Muralidharan
38516
Not able to understand please tell some other method.
– priyanka kumari
Dec 2 at 5:31
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
From
begin{align*}
(1-x)^n = sum (-1)^i binom{n}{i} x^i
end{align*}
differentiating, we obtain
begin{align*}
n(1-x)^{n-1} = sum (-1)^ii binom{n}{i} x^{i-1}
end{align*}
Multiply by $x$ and differentiate:
begin{align*}
n(1-x)^{n-1}x &= sum (-1)^ii binom{n}{i} x^{i}\
n(n-1)(1-x)^{n-2}x + n(1-x)^{n-1} = sum (-1)^i i^2 binom{n}{i} x^{i-1}
end{align*}
Use the same technique for obtaining an expression for $sum (-1)^i i^3 x^i binom{n}{i}$.
The given expression can be expanded as
begin{align*}
sum (-1)^i binom{n}{i} [abc - i(a+b+c) + i^2(ab+bc+ca) -i^3]
end{align*}
Put $x=1$ and see that all the individual expressions evaluate to 0.
answered Dec 2 at 4:23
Muralidharan
38516
From
begin{align*}
(1-x)^n = sum (-1)^i binom{n}{i} x^i
end{align*}
differentiating, we obtain
begin{align*}
n(1-x)^{n-1} = sum (-1)^ii binom{n}{i} x^{i-1}
end{align*}
Multiply by $x$ and differentiate:
begin{align*}
n(1-x)^{n-1}x &= sum (-1)^ii binom{n}{i} x^{i}\
n(n-1)(1-x)^{n-2}x + n(1-x)^{n-1} = sum (-1)^i i^2 binom{n}{i} x^{i-1}
end{align*}
Use the same technique for obtaining an expression for $sum (-1)^i i^3 x^i binom{n}{i}$.
The given expression can be expanded as
begin{align*}
sum (-1)^i binom{n}{i} [abc - i(a+b+c) + i^2(ab+bc+ca) -i^3]
end{align*}
Put $x=1$ and see that all the individual expressions evaluate to 0.
answered Dec 2 at 4:23
Muralidharan
38516
edited Dec 2 at 4:32
answered Dec 2 at 4:23
Muralidharan
38516
answered Dec 2 at 4:23
Muralidharan
38516
answered Dec 2 at 4:23
Muralidharan
38516
38516
Not able to understand please tell some other method.
– priyanka kumari
Dec 2 at 5:31
add a comment |
Not able to understand please tell some other method.
– priyanka kumari
Dec 2 at 5:31
Not able to understand please tell some other method.
– priyanka kumari
Dec 2 at 5:31
Not able to understand please tell some other method.
– priyanka kumari
Dec 2 at 5:31
add a comment |
up vote
0
down vote
$(x-1)^n=C_0x^n-C_1x^{n-1}+C_2x^{n-2}...+(-1)^nC_n\$
Multiply by $x^{a-n}$,
$implies (x-1)^nx^{a-n}=C_0x^a-C_1x^{a-1}+C_2x^{a-2}...+(-1)^nC_nx^{a-n}\$
Differentiate with respect to $x$,
$n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}=aC_0x^{a-1}-(a-1)C_1x^{a-2}+(a-2)C_2x^{a-3}...+(-1)^n(a-n)C_nx^{a-n-1}\$
Multiply with $x^{b-a+1}$,
$implies x^{b-a+1}[n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}]=n(x-1)^{n-1}x^{b-n+1}+(a-n)(x-1)^nx^{b-n}=aC_0x^b-(a-1)C_1x^{b-1}+(a-2)C_2x^{b-2}...+(-1)^n(a-n)C_nx^{b-n}\$
Differentiate with respect to $x$,
$n(n-1)(x-1)^{n-2}x^{b-n+1}+n(b-n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}+n(a-n)(x-1)^{n-1}x^{b-n}=abC_0x^{b-1}-(a-1)(b-1)C_1x^{b-2}+(a-2)(b-2)C_2x^{b-3}...+(-1)^n(a-n)(b-n)C_nx^{b-n-1}\$
Multiply with $x^{c-b+1}$,
$x^{c-b+1}[n(n-1)(x-1)^{n-2}x^{b-n+1}+n(a+b-2n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}]=n(n-1)(x-1)^{n-2}x^{c-n+2}+n(a+b-2n+1)(x-1)^{n-1}x^{c-n+1}+(a-n)(b-n)(x-1)^nx^{c-n}=abC_0x^c-(a-1)(b-1)C_1x^{c-1}+(a-2)(b-2)C_2x^{c-2}...+(-1)^n(a-n)(b-n)C_nx^{c-n}\$
Differentiate with respect to $x$,
$n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(c-n+2)(x-1)^{n-2}x^{c-n+1}+n(n-1)(a+b-2n+1)(x-1)^{n-2}x^{c-n+1}+n(c-n+1)(a+b-2n+1)(x-1)^{n-1}x^{c-n}+n(a-n)(b-n)(x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(a+b+c-3n+3)(x-1)^{n-2}x^{c-n+1}+n[(c-n+1)(a+b-2n+1)+(a-n)(b-n)](x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=abcC_0x^{c-1}-(a-1)(b-1)(c-1)C_1x^{c-2}+(a-2)(b-2)(c-2)C_2x^{c-3}...+(-1)^n(a-n)(b-n)(c-n)C_nx^{c-n-1}\$
Set $n>3, x=1$,
$0=abcC_0-(a-1)(b-1)(c-1)C_1+(a-2)(b-2)(c-2)C_2...+(-1)^n(a-n)(b-n)(c-n)C_n\$
Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4, a=b=c=1/2$, and it seems it should be $abcC_0$.
answered Dec 2 at 6:46
Shubham Johri
1,08339
add a comment |
up vote
0
down vote
$(x-1)^n=C_0x^n-C_1x^{n-1}+C_2x^{n-2}...+(-1)^nC_n\$
Multiply by $x^{a-n}$,
$implies (x-1)^nx^{a-n}=C_0x^a-C_1x^{a-1}+C_2x^{a-2}...+(-1)^nC_nx^{a-n}\$
Differentiate with respect to $x$,
$n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}=aC_0x^{a-1}-(a-1)C_1x^{a-2}+(a-2)C_2x^{a-3}...+(-1)^n(a-n)C_nx^{a-n-1}\$
Multiply with $x^{b-a+1}$,
$implies x^{b-a+1}[n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}]=n(x-1)^{n-1}x^{b-n+1}+(a-n)(x-1)^nx^{b-n}=aC_0x^b-(a-1)C_1x^{b-1}+(a-2)C_2x^{b-2}...+(-1)^n(a-n)C_nx^{b-n}\$
Differentiate with respect to $x$,
$n(n-1)(x-1)^{n-2}x^{b-n+1}+n(b-n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}+n(a-n)(x-1)^{n-1}x^{b-n}=abC_0x^{b-1}-(a-1)(b-1)C_1x^{b-2}+(a-2)(b-2)C_2x^{b-3}...+(-1)^n(a-n)(b-n)C_nx^{b-n-1}\$
Multiply with $x^{c-b+1}$,
$x^{c-b+1}[n(n-1)(x-1)^{n-2}x^{b-n+1}+n(a+b-2n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}]=n(n-1)(x-1)^{n-2}x^{c-n+2}+n(a+b-2n+1)(x-1)^{n-1}x^{c-n+1}+(a-n)(b-n)(x-1)^nx^{c-n}=abC_0x^c-(a-1)(b-1)C_1x^{c-1}+(a-2)(b-2)C_2x^{c-2}...+(-1)^n(a-n)(b-n)C_nx^{c-n}\$
Differentiate with respect to $x$,
$n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(c-n+2)(x-1)^{n-2}x^{c-n+1}+n(n-1)(a+b-2n+1)(x-1)^{n-2}x^{c-n+1}+n(c-n+1)(a+b-2n+1)(x-1)^{n-1}x^{c-n}+n(a-n)(b-n)(x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(a+b+c-3n+3)(x-1)^{n-2}x^{c-n+1}+n[(c-n+1)(a+b-2n+1)+(a-n)(b-n)](x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=abcC_0x^{c-1}-(a-1)(b-1)(c-1)C_1x^{c-2}+(a-2)(b-2)(c-2)C_2x^{c-3}...+(-1)^n(a-n)(b-n)(c-n)C_nx^{c-n-1}\$
Set $n>3, x=1$,
$0=abcC_0-(a-1)(b-1)(c-1)C_1+(a-2)(b-2)(c-2)C_2...+(-1)^n(a-n)(b-n)(c-n)C_n\$
Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4, a=b=c=1/2$, and it seems it should be $abcC_0$.
answered Dec 2 at 6:46
Shubham Johri
1,08339
add a comment |
up vote
0
down vote
up vote
0
down vote
$(x-1)^n=C_0x^n-C_1x^{n-1}+C_2x^{n-2}...+(-1)^nC_n\$
Multiply by $x^{a-n}$,
$implies (x-1)^nx^{a-n}=C_0x^a-C_1x^{a-1}+C_2x^{a-2}...+(-1)^nC_nx^{a-n}\$
Differentiate with respect to $x$,
$n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}=aC_0x^{a-1}-(a-1)C_1x^{a-2}+(a-2)C_2x^{a-3}...+(-1)^n(a-n)C_nx^{a-n-1}\$
Multiply with $x^{b-a+1}$,
$implies x^{b-a+1}[n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}]=n(x-1)^{n-1}x^{b-n+1}+(a-n)(x-1)^nx^{b-n}=aC_0x^b-(a-1)C_1x^{b-1}+(a-2)C_2x^{b-2}...+(-1)^n(a-n)C_nx^{b-n}\$
Differentiate with respect to $x$,
$n(n-1)(x-1)^{n-2}x^{b-n+1}+n(b-n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}+n(a-n)(x-1)^{n-1}x^{b-n}=abC_0x^{b-1}-(a-1)(b-1)C_1x^{b-2}+(a-2)(b-2)C_2x^{b-3}...+(-1)^n(a-n)(b-n)C_nx^{b-n-1}\$
Multiply with $x^{c-b+1}$,
$x^{c-b+1}[n(n-1)(x-1)^{n-2}x^{b-n+1}+n(a+b-2n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}]=n(n-1)(x-1)^{n-2}x^{c-n+2}+n(a+b-2n+1)(x-1)^{n-1}x^{c-n+1}+(a-n)(b-n)(x-1)^nx^{c-n}=abC_0x^c-(a-1)(b-1)C_1x^{c-1}+(a-2)(b-2)C_2x^{c-2}...+(-1)^n(a-n)(b-n)C_nx^{c-n}\$
Differentiate with respect to $x$,
$n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(c-n+2)(x-1)^{n-2}x^{c-n+1}+n(n-1)(a+b-2n+1)(x-1)^{n-2}x^{c-n+1}+n(c-n+1)(a+b-2n+1)(x-1)^{n-1}x^{c-n}+n(a-n)(b-n)(x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(a+b+c-3n+3)(x-1)^{n-2}x^{c-n+1}+n[(c-n+1)(a+b-2n+1)+(a-n)(b-n)](x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=abcC_0x^{c-1}-(a-1)(b-1)(c-1)C_1x^{c-2}+(a-2)(b-2)(c-2)C_2x^{c-3}...+(-1)^n(a-n)(b-n)(c-n)C_nx^{c-n-1}\$
Set $n>3, x=1$,
$0=abcC_0-(a-1)(b-1)(c-1)C_1+(a-2)(b-2)(c-2)C_2...+(-1)^n(a-n)(b-n)(c-n)C_n\$
Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4, a=b=c=1/2$, and it seems it should be $abcC_0$.
answered Dec 2 at 6:46
Shubham Johri
1,08339
$(x-1)^n=C_0x^n-C_1x^{n-1}+C_2x^{n-2}...+(-1)^nC_n\$
Multiply by $x^{a-n}$,
$implies (x-1)^nx^{a-n}=C_0x^a-C_1x^{a-1}+C_2x^{a-2}...+(-1)^nC_nx^{a-n}\$
Differentiate with respect to $x$,
$n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}=aC_0x^{a-1}-(a-1)C_1x^{a-2}+(a-2)C_2x^{a-3}...+(-1)^n(a-n)C_nx^{a-n-1}\$
Multiply with $x^{b-a+1}$,
$implies x^{b-a+1}[n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}]=n(x-1)^{n-1}x^{b-n+1}+(a-n)(x-1)^nx^{b-n}=aC_0x^b-(a-1)C_1x^{b-1}+(a-2)C_2x^{b-2}...+(-1)^n(a-n)C_nx^{b-n}\$
Differentiate with respect to $x$,
$n(n-1)(x-1)^{n-2}x^{b-n+1}+n(b-n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}+n(a-n)(x-1)^{n-1}x^{b-n}=abC_0x^{b-1}-(a-1)(b-1)C_1x^{b-2}+(a-2)(b-2)C_2x^{b-3}...+(-1)^n(a-n)(b-n)C_nx^{b-n-1}\$
Multiply with $x^{c-b+1}$,
$x^{c-b+1}[n(n-1)(x-1)^{n-2}x^{b-n+1}+n(a+b-2n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}]=n(n-1)(x-1)^{n-2}x^{c-n+2}+n(a+b-2n+1)(x-1)^{n-1}x^{c-n+1}+(a-n)(b-n)(x-1)^nx^{c-n}=abC_0x^c-(a-1)(b-1)C_1x^{c-1}+(a-2)(b-2)C_2x^{c-2}...+(-1)^n(a-n)(b-n)C_nx^{c-n}\$
Differentiate with respect to $x$,
$n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(c-n+2)(x-1)^{n-2}x^{c-n+1}+n(n-1)(a+b-2n+1)(x-1)^{n-2}x^{c-n+1}+n(c-n+1)(a+b-2n+1)(x-1)^{n-1}x^{c-n}+n(a-n)(b-n)(x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(a+b+c-3n+3)(x-1)^{n-2}x^{c-n+1}+n[(c-n+1)(a+b-2n+1)+(a-n)(b-n)](x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=abcC_0x^{c-1}-(a-1)(b-1)(c-1)C_1x^{c-2}+(a-2)(b-2)(c-2)C_2x^{c-3}...+(-1)^n(a-n)(b-n)(c-n)C_nx^{c-n-1}\$
Set $n>3, x=1$,
$0=abcC_0-(a-1)(b-1)(c-1)C_1+(a-2)(b-2)(c-2)C_2...+(-1)^n(a-n)(b-n)(c-n)C_n\$
Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4, a=b=c=1/2$, and it seems it should be $abcC_0$.
answered Dec 2 at 6:46
Shubham Johri
1,08339
edited Dec 2 at 14:13
answered Dec 2 at 6:46
Shubham Johri
1,08339
answered Dec 2 at 6:46
Shubham Johri
1,08339
answered Dec 2 at 6:46
Shubham Johri
1,08339
1,08339
add a comment |
add a comment |
What are the $C_k$ ? What are $a,b,c$?
– darij grinberg
Dec 2 at 3:51
$C_k$ is the binomial coefficient $binom nk$
– Shubham Johri
Dec 2 at 5:23
@priyanka kumari Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4,a=b=c=1/2$, and it seems it should be $abcC_0$.
– Shubham Johri
Dec 2 at 6:56