My Math Forum Problem in limits of summations

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 July 23rd, 2011, 11:20 AM #1 Newbie   Joined: Jul 2011 Posts: 4 Thanks: 0 Problem in limits of summations I have this expression: $\sum_{n=1}^\infty\sum_{a=0}^{n}\sum_{c=0}^{n-a}\sum_{d=0}^{a}\sum_{k=0}^{c+d}\frac{1}{n}\binom{ n}{a}\binom{n+a}{n-1}\binom{n-a}{c}\binom{a}{d}\binom{c+d}{k}(-1)^{k}x^{n+a+c+d}y^{c+a-k}$. I am going to set $n+a+c+d=m$ and $c+a-k=r$, so I have to change summations' limits I know that the first summation will be: $\sum_{m=1}^\infty$ because ${n \rightarrow \infty}$ and $a,c,d\ge 0$. and that n should not appear. But that i can't understand is what happens to limits of the other summations ${a \le ?}$ and ${c \le ?}$ and why. I know that i have to use these inequalities: $0\le a\le n, 0\le c \le n-a, 0 \le d \le a, 0\le k \le c+d$. Any idea?
 July 23rd, 2011, 04:23 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,947 Thanks: 1139 Math Focus: Elementary mathematics and beyond Re: Problem in limits of summations There is a problem with $m$. If you swap $n$ with $m$ you end up with $m\,=\,m\,+\,a\,+\,c\,+\,d$, which isn't going to work.
 July 23rd, 2011, 04:43 PM #3 Newbie   Joined: Jul 2011 Posts: 4 Thanks: 0 Re: Problem in limits of summations I am not going to swap $m$ with $n$. My final goal is the expression with summations to end up like that: $x^{m}$ and $y^{r}$ so what about the limits? Thank you for your reply and I am grateful to help me!
 July 23rd, 2011, 04:47 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,947 Thanks: 1139 Math Focus: Elementary mathematics and beyond Re: Problem in limits of summations You've got the assignment $m\,=\,n\,+\,a\,+\,c\,+\,d$ Then you state that you are going to use $\sum_{m=1}^{\infty}$ What happens to $n$ ?
 July 23rd, 2011, 04:57 PM #5 Newbie   Joined: Jul 2011 Posts: 4 Thanks: 0 Re: Problem in limits of summations As I explained because of: $n$ takes numbers from 1 until $\infty$ and $a,c,d\ge 0$ then $m=n+a+c+d$ will take numbers from 1 until $\infty$.

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