Re: proof by induction?

*To*: mathgroup at smc.vnet.net*Subject*: [mg29741] Re: [mg29729] proof by induction?*From*: Tomas Garza <tgarza01 at prodigy.net.mx>*Date*: Fri, 6 Jul 2001 03:24:36 -0400 (EDT)*References*: <200107042243.SAA17412@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

I wonder what it is you're talking about, but my guess is that you want to calculate the sum of the first n + 1 terms of a geometric series with ratio r and a constant coefficient A, and prove the result by induction (which is just an exercise in proving by induction, since you can derive the formula in a straightforward way). If such is the case, you have a syntax error in your formula, which should read A(r^(n+1)-1)/(r-1) instead of A((r^n+1)-1)/(r-1), and, second, the problem has nothing to do with Mathematica, although you could use it in a most inefficient way it to obtain the sum of two terms. See, actually you are saying that, if the formula is true for n = 1 (which is obvious), and if from the assumption that it is true for a given n it can be proved that it is true for n + 1, then it is true for any finite n. In Mathematica you would define In[1]:= (*Assume that the following is true, i..e., the sum of the first n terms of the geometric series is *) s[n_] := (-1 + r^n)/(-1 + r) Then, adding one more term, i.e., r^n gives s[n] + r^n: In[2]:= Simplify[s[n] + r^n] Out[2]= (-1 + r^(1 + n))/(-1 + r) which is of course s[n + 1]: In[3]:= Simplify[s[n] + r^n] == s[n + 1] Out[3]= True Q.E.D. But this is trying to kill a mosquito with a cannon. Tomas Garza Mexico City ----- Original Message ----- From: <WebMaster at FarOffGrace.com> To: mathgroup at smc.vnet.net Subject: [mg29741] [mg29729] proof by induction? > > I'm trying to solve this question, where I must prove this equation by > induction, but im totally stumped, im hoping you can help me out, thanks > the equation is > > S= A((r^n+1)-1)/(r-1) > >

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