Explanation for binomial sums $sum_{n=0}^{infty} binom{-4}{n-1} (-1)^{n-1} x^n = sum_{n=0}^{infty}...
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I was looking at some negative binomial coefficient problems and I stumbled upon this explanation
$$sum_{n=0}^{infty} binom{n+2}{3} x^n = sum_{n=0}^{infty} binom{n+2}{n-1} x^n= sum_{n=0}^{infty} binom{-4}{n-1} (-1)^{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n} (-1) x^{n+1} $$
I was wondering how the author arrived at the conclusion that
$$sum_{n=0}^{infty} binom{n+2}{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n-1} (-1)^{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n} (-1) x^{n+1}$$
combinatorics binomial-coefficients negative-binomial
asked Nov 13 at 20:42
Itsnhantransitive
769315
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up vote
1
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I was looking at some negative binomial coefficient problems and I stumbled upon this explanation
$$sum_{n=0}^{infty} binom{n+2}{3} x^n = sum_{n=0}^{infty} binom{n+2}{n-1} x^n= sum_{n=0}^{infty} binom{-4}{n-1} (-1)^{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n} (-1) x^{n+1} $$
I was wondering how the author arrived at the conclusion that
$$sum_{n=0}^{infty} binom{n+2}{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n-1} (-1)^{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n} (-1) x^{n+1}$$
combinatorics binomial-coefficients negative-binomial
asked Nov 13 at 20:42
Itsnhantransitive
769315
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was looking at some negative binomial coefficient problems and I stumbled upon this explanation
$$sum_{n=0}^{infty} binom{n+2}{3} x^n = sum_{n=0}^{infty} binom{n+2}{n-1} x^n= sum_{n=0}^{infty} binom{-4}{n-1} (-1)^{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n} (-1) x^{n+1} $$
I was wondering how the author arrived at the conclusion that
$$sum_{n=0}^{infty} binom{n+2}{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n-1} (-1)^{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n} (-1) x^{n+1}$$
combinatorics binomial-coefficients negative-binomial
asked Nov 13 at 20:42
Itsnhantransitive
769315
I was looking at some negative binomial coefficient problems and I stumbled upon this explanation
$$sum_{n=0}^{infty} binom{n+2}{3} x^n = sum_{n=0}^{infty} binom{n+2}{n-1} x^n= sum_{n=0}^{infty} binom{-4}{n-1} (-1)^{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n} (-1) x^{n+1} $$
I was wondering how the author arrived at the conclusion that
$$sum_{n=0}^{infty} binom{n+2}{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n-1} (-1)^{n-1} x^n = sum_{n=0}^{infty} binom{-4}{n} (-1) x^{n+1}$$
combinatorics binomial-coefficients negative-binomial
combinatorics binomial-coefficients negative-binomial
asked Nov 13 at 20:42
Itsnhantransitive
769315
asked Nov 13 at 20:42
Itsnhantransitive
769315
asked Nov 13 at 20:42
Itsnhantransitive
769315
asked Nov 13 at 20:42
Itsnhantransitive
769315
asked Nov 13 at 20:42
Itsnhantransitive
769315
769315
add a comment |
add a comment |
1 Answer
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Index transformation (replace $n$ by $n+1$) gives
$$ sum_{n=0}^infty binom{-4}{n-1}(-1)^{n-1} x^n
= sum_{n=-1}^infty binom{-4}{n}(-1)^n x^{n+1} $$
Now, on the right hand side, the term for $n=-1$, namely
$$ binom{-4}{-1} (-1)^{-1} x^0 = 0 $$
is zero due to $binom{-4}{-1} = 0$. Hence
$$ sum_{n=0}^infty binom{-4}{n-1}(-1)^{n-1} x^n
= sum_{n=-1}^infty binom{-4}{n}(-1)^n x^{n+1}
= sum_{n=0}^infty binom{-4}{n}(-1)^n x^{n+1} $$
answered Nov 13 at 20:46
martini
70k45990
ah ok I see! Thank you!
– Itsnhantransitive
Nov 13 at 21:26
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Index transformation (replace $n$ by $n+1$) gives
$$ sum_{n=0}^infty binom{-4}{n-1}(-1)^{n-1} x^n
= sum_{n=-1}^infty binom{-4}{n}(-1)^n x^{n+1} $$
Now, on the right hand side, the term for $n=-1$, namely
$$ binom{-4}{-1} (-1)^{-1} x^0 = 0 $$
is zero due to $binom{-4}{-1} = 0$. Hence
$$ sum_{n=0}^infty binom{-4}{n-1}(-1)^{n-1} x^n
= sum_{n=-1}^infty binom{-4}{n}(-1)^n x^{n+1}
= sum_{n=0}^infty binom{-4}{n}(-1)^n x^{n+1} $$
answered Nov 13 at 20:46
martini
70k45990
ah ok I see! Thank you!
– Itsnhantransitive
Nov 13 at 21:26
add a comment |
up vote
1
down vote
accepted
Index transformation (replace $n$ by $n+1$) gives
$$ sum_{n=0}^infty binom{-4}{n-1}(-1)^{n-1} x^n
= sum_{n=-1}^infty binom{-4}{n}(-1)^n x^{n+1} $$
Now, on the right hand side, the term for $n=-1$, namely
$$ binom{-4}{-1} (-1)^{-1} x^0 = 0 $$
is zero due to $binom{-4}{-1} = 0$. Hence
$$ sum_{n=0}^infty binom{-4}{n-1}(-1)^{n-1} x^n
= sum_{n=-1}^infty binom{-4}{n}(-1)^n x^{n+1}
= sum_{n=0}^infty binom{-4}{n}(-1)^n x^{n+1} $$
answered Nov 13 at 20:46
martini
70k45990
ah ok I see! Thank you!
– Itsnhantransitive
Nov 13 at 21:26
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Index transformation (replace $n$ by $n+1$) gives
$$ sum_{n=0}^infty binom{-4}{n-1}(-1)^{n-1} x^n
= sum_{n=-1}^infty binom{-4}{n}(-1)^n x^{n+1} $$
Now, on the right hand side, the term for $n=-1$, namely
$$ binom{-4}{-1} (-1)^{-1} x^0 = 0 $$
is zero due to $binom{-4}{-1} = 0$. Hence
$$ sum_{n=0}^infty binom{-4}{n-1}(-1)^{n-1} x^n
= sum_{n=-1}^infty binom{-4}{n}(-1)^n x^{n+1}
= sum_{n=0}^infty binom{-4}{n}(-1)^n x^{n+1} $$
answered Nov 13 at 20:46
martini
70k45990
Index transformation (replace $n$ by $n+1$) gives
$$ sum_{n=0}^infty binom{-4}{n-1}(-1)^{n-1} x^n
= sum_{n=-1}^infty binom{-4}{n}(-1)^n x^{n+1} $$
Now, on the right hand side, the term for $n=-1$, namely
$$ binom{-4}{-1} (-1)^{-1} x^0 = 0 $$
is zero due to $binom{-4}{-1} = 0$. Hence
$$ sum_{n=0}^infty binom{-4}{n-1}(-1)^{n-1} x^n
= sum_{n=-1}^infty binom{-4}{n}(-1)^n x^{n+1}
= sum_{n=0}^infty binom{-4}{n}(-1)^n x^{n+1} $$
answered Nov 13 at 20:46
martini
70k45990
answered Nov 13 at 20:46
martini
70k45990
answered Nov 13 at 20:46
martini
70k45990
answered Nov 13 at 20:46
martini
70k45990
70k45990
ah ok I see! Thank you!
– Itsnhantransitive
Nov 13 at 21:26
add a comment |
ah ok I see! Thank you!
– Itsnhantransitive
Nov 13 at 21:26
ah ok I see! Thank you!
– Itsnhantransitive
Nov 13 at 21:26
ah ok I see! Thank you!
– Itsnhantransitive
Nov 13 at 21:26
add a comment |
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