Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!swrinde!zaphod.mps.ohio-state.edu!cis.ohio-state.edu!ucbvax!fshvmfk1.vnet.ibm.com!raghu From: raghu@fshvmfk1.vnet.ibm.com Newsgroups: comp.theory Subject: Re: expression for F(n) = a**0 + a**1 + ... + a**n Message-ID: <9106190652.AA05711@ucbvax.Berkeley.EDU> Date: 18 Jun 91 12:16:52 GMT Sender: daemon@ucbvax.BERKELEY.EDU Lines: 31 >Subject: expression for F(n) = a**0 + a**1 + ... + a**(n-1) + a**n >From: ping@bnlux1.bnl.gov (Shiping Zhang) >Message-ID: <1991Jun18.022651.16732@bnlux1.bnl.gov> >Date: 18 Jun 91 02:26:51 GMT > >I'm wondering if there is an expression for F(n) = a**0 + a**1 + ... +a**n >in term of both a and n. Here a is a constant. I know the expression fr >a = 2, which is 2**(n+1) -1. But now I want to know the expression fo >any constant a. Thanks for any help. > >-ping F(n) is a geometric series, where the ith term is given by the the recurrence relation : T(i) = r*T(i-1) r is called the common ratio. The sum of n terms is T(0) * (r**n - 1) / (r - 1) In your case T(0) = 1, r = a, n = n+1, so the sum is (r**(n+1) - 1) / (r -1). If r = a = 2, the sum is is the same as the one you have. Raghu V. Hudli IBM Corp ====================================================================== Disclaimer: The opinions/views expressed in this note are my own and do not necessarily reflect those of my employer.