Jun 6, 2005 #1 phonic Messages 28 Reaction score 0 Is it possible to get a analytical result for this series? It looks simple: \sum_{k=1} ^t a^{t-k}b^{k-1} Thanks a lot!
Is it possible to get a analytical result for this series? It looks simple: \sum_{k=1} ^t a^{t-k}b^{k-1} Thanks a lot!
Jun 6, 2005 #2 dextercioby Science Advisor Insights Author Messages 13,394 Reaction score 4,064 Hmm.Use [ tex ] & [ /tex ] commands (without the spaces) for opening & closing tex tags. That's no series,it's a polynomial in 2 variables. Daniel.
Hmm.Use [ tex ] & [ /tex ] commands (without the spaces) for opening & closing tex tags. That's no series,it's a polynomial in 2 variables. Daniel.
Jun 6, 2005 #3 phonic Messages 28 Reaction score 0 Thanks for your corection. It's my first time to post message here. Yes, this is a polynomial, with all coeficient as 1. Is there some method to deal with it?
Thanks for your corection. It's my first time to post message here. Yes, this is a polynomial, with all coeficient as 1. Is there some method to deal with it?
Jun 6, 2005 #4 dextercioby Science Advisor Insights Author Messages 13,394 Reaction score 4,064 Yes,try to write some terms in the sum and then see whether you recognize something familiar. Daniel.
Yes,try to write some terms in the sum and then see whether you recognize something familiar. Daniel.
Jun 6, 2005 #5 phonic Messages 28 Reaction score 0 Thanks for your hints. I think it can be calculated in this way: \sum_{k=1} ^t a^{t-k}b^{k-1} = \frac{a^t}{b}\sum_{k=1} ^t (\frac{b}{a})^{k}
Thanks for your hints. I think it can be calculated in this way: \sum_{k=1} ^t a^{t-k}b^{k-1} = \frac{a^t}{b}\sum_{k=1} ^t (\frac{b}{a})^{k}
Jun 6, 2005 #6 dextercioby Science Advisor Insights Author Messages 13,394 Reaction score 4,064 It's easier this way: \sum_{k=1}^{t} a^{t-k}b^{k-1}=a^{t-1}b^{0}+a^{t-2}b^{1}+...+a^{1}b^{t-2}+a^{0}b^{t-1}=\frac{a^{t}-b^{t}}{a-b} with "t" uneven. Daniel.
It's easier this way: \sum_{k=1}^{t} a^{t-k}b^{k-1}=a^{t-1}b^{0}+a^{t-2}b^{1}+...+a^{1}b^{t-2}+a^{0}b^{t-1}=\frac{a^{t}-b^{t}}{a-b} with "t" uneven. Daniel.