Coefficient of $x^r$ in $frac{1}{(1-ax)^n}$
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In my combinatorics class, we were taught that the coefficient of $x^r$ in $frac{1}{(1-x)^n}$ is $binom{r+n-1}{r}$. I am struggling to figure out what the coefficient of $x^r$ would be in $frac{1}{(1-ax)^n}$, where $a in mathbb{N}$. I know that the expansion of $frac{1}{(1-x)^n}$ is $sumlimits_{k=0}^infty binom{k+n-1}{k}x^k$, so my initial thought was that it would be $a^rbinom{r+n-1}{r}$, but I'm not sure. Any help would be appreciated.
combinatorics generating-functions
asked Oct 19 '17 at 21:46
wjmcalliwjmcalli
5115
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add a comment |
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In my combinatorics class, we were taught that the coefficient of $x^r$ in $frac{1}{(1-x)^n}$ is $binom{r+n-1}{r}$. I am struggling to figure out what the coefficient of $x^r$ would be in $frac{1}{(1-ax)^n}$, where $a in mathbb{N}$. I know that the expansion of $frac{1}{(1-x)^n}$ is $sumlimits_{k=0}^infty binom{k+n-1}{k}x^k$, so my initial thought was that it would be $a^rbinom{r+n-1}{r}$, but I'm not sure. Any help would be appreciated.
combinatorics generating-functions
asked Oct 19 '17 at 21:46
wjmcalliwjmcalli
5115
$endgroup$
2
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Put $y = ax$. Then you will have the term $binom{r + n - 1}{k}y^r = binom{r + n - 1}{k}(ax)^r = a^rbinom{r + n - 1}{k}x^r$.
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– Abhiram Natarajan
Oct 19 '17 at 21:49
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What is $k$ w.r.t. $n$ and $r$?
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– Bernard
Oct 19 '17 at 22:11
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Should have said $r$, my mistake.
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– wjmcalli
Oct 19 '17 at 22:13
add a comment |
$begingroup$
In my combinatorics class, we were taught that the coefficient of $x^r$ in $frac{1}{(1-x)^n}$ is $binom{r+n-1}{r}$. I am struggling to figure out what the coefficient of $x^r$ would be in $frac{1}{(1-ax)^n}$, where $a in mathbb{N}$. I know that the expansion of $frac{1}{(1-x)^n}$ is $sumlimits_{k=0}^infty binom{k+n-1}{k}x^k$, so my initial thought was that it would be $a^rbinom{r+n-1}{r}$, but I'm not sure. Any help would be appreciated.
combinatorics generating-functions
asked Oct 19 '17 at 21:46
wjmcalliwjmcalli
5115
$endgroup$
In my combinatorics class, we were taught that the coefficient of $x^r$ in $frac{1}{(1-x)^n}$ is $binom{r+n-1}{r}$. I am struggling to figure out what the coefficient of $x^r$ would be in $frac{1}{(1-ax)^n}$, where $a in mathbb{N}$. I know that the expansion of $frac{1}{(1-x)^n}$ is $sumlimits_{k=0}^infty binom{k+n-1}{k}x^k$, so my initial thought was that it would be $a^rbinom{r+n-1}{r}$, but I'm not sure. Any help would be appreciated.
combinatorics generating-functions
combinatorics generating-functions
asked Oct 19 '17 at 21:46
wjmcalliwjmcalli
5115
asked Oct 19 '17 at 21:46
wjmcalliwjmcalli
5115
edited Dec 7 '18 at 20:48
wjmcalli
asked Oct 19 '17 at 21:46
wjmcalliwjmcalli
5115
asked Oct 19 '17 at 21:46
wjmcalliwjmcalli
5115
asked Oct 19 '17 at 21:46
wjmcalliwjmcalli
5115
5115
2
$begingroup$
Put $y = ax$. Then you will have the term $binom{r + n - 1}{k}y^r = binom{r + n - 1}{k}(ax)^r = a^rbinom{r + n - 1}{k}x^r$.
$endgroup$
– Abhiram Natarajan
Oct 19 '17 at 21:49
$begingroup$
What is $k$ w.r.t. $n$ and $r$?
$endgroup$
– Bernard
Oct 19 '17 at 22:11
$begingroup$
Should have said $r$, my mistake.
$endgroup$
– wjmcalli
Oct 19 '17 at 22:13
add a comment |
2
$begingroup$
Put $y = ax$. Then you will have the term $binom{r + n - 1}{k}y^r = binom{r + n - 1}{k}(ax)^r = a^rbinom{r + n - 1}{k}x^r$.
$endgroup$
– Abhiram Natarajan
Oct 19 '17 at 21:49
$begingroup$
What is $k$ w.r.t. $n$ and $r$?
$endgroup$
– Bernard
Oct 19 '17 at 22:11
$begingroup$
Should have said $r$, my mistake.
$endgroup$
– wjmcalli
Oct 19 '17 at 22:13
2
2
$begingroup$
Put $y = ax$. Then you will have the term $binom{r + n - 1}{k}y^r = binom{r + n - 1}{k}(ax)^r = a^rbinom{r + n - 1}{k}x^r$.
$endgroup$
– Abhiram Natarajan
Oct 19 '17 at 21:49
$begingroup$
Put $y = ax$. Then you will have the term $binom{r + n - 1}{k}y^r = binom{r + n - 1}{k}(ax)^r = a^rbinom{r + n - 1}{k}x^r$.
$endgroup$
– Abhiram Natarajan
Oct 19 '17 at 21:49
$begingroup$
What is $k$ w.r.t. $n$ and $r$?
$endgroup$
– Bernard
Oct 19 '17 at 22:11
$begingroup$
What is $k$ w.r.t. $n$ and $r$?
$endgroup$
– Bernard
Oct 19 '17 at 22:11
$begingroup$
Should have said $r$, my mistake.
$endgroup$
– wjmcalli
Oct 19 '17 at 22:13
$begingroup$
Should have said $r$, my mistake.
$endgroup$
– wjmcalli
Oct 19 '17 at 22:13
add a comment |
1 Answer
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Hint: See that coefficient of $x^r$ will be $a^rbinom{r + n - 1}{r}$.
answered Oct 20 '17 at 0:30
user8795user8795
5,68462047
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$begingroup$
Hint: See that coefficient of $x^r$ will be $a^rbinom{r + n - 1}{r}$.
answered Oct 20 '17 at 0:30
user8795user8795
5,68462047
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add a comment |
$begingroup$
Hint: See that coefficient of $x^r$ will be $a^rbinom{r + n - 1}{r}$.
answered Oct 20 '17 at 0:30
user8795user8795
5,68462047
$endgroup$
add a comment |
$begingroup$
Hint: See that coefficient of $x^r$ will be $a^rbinom{r + n - 1}{r}$.
answered Oct 20 '17 at 0:30
user8795user8795
5,68462047
$endgroup$
Hint: See that coefficient of $x^r$ will be $a^rbinom{r + n - 1}{r}$.
answered Oct 20 '17 at 0:30
user8795user8795
5,68462047
answered Oct 20 '17 at 0:30
user8795user8795
5,68462047
answered Oct 20 '17 at 0:30
user8795user8795
5,68462047
answered Oct 20 '17 at 0:30
user8795user8795
5,68462047
5,68462047
add a comment |
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2
$begingroup$
Put $y = ax$. Then you will have the term $binom{r + n - 1}{k}y^r = binom{r + n - 1}{k}(ax)^r = a^rbinom{r + n - 1}{k}x^r$.
$endgroup$
– Abhiram Natarajan
Oct 19 '17 at 21:49
$begingroup$
What is $k$ w.r.t. $n$ and $r$?
$endgroup$
– Bernard
Oct 19 '17 at 22:11
$begingroup$
Should have said $r$, my mistake.
$endgroup$
– wjmcalli
Oct 19 '17 at 22:13