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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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