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Induction Proof

Date: 10/15/97 at 01:56:48
From: Masika bryce
Subject: Induction proof.

This induction problem asks us to find the formula by examining the 
values of the expression  1/1.2 + 1/2.3 + .... + 1/n(n+1), and to 
prove the result.

I have set up a formula: => 2^n-1/n(n+1) and have tried to prove it is 
true.  I could not continue because I did not really know how to 
substitute for P(n+1). Please help.

Date: 10/15/97 at 08:37:14
From: Doctor Jerry
Subject: Re: Induction proof.

Hi Masika,

I think your formula is not correct.  Here's a way of finding a 
correct formula in this case.

First, notice that 1/(n(n+1)) = 1/n - 1/(n+1).  Then,

1/(1.2)+1/(2.3)+...+1/(n(n+1)) = 1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)

You can see that there are many cancellations, leaving

1/(1.2)+1/(2.3)+...+1/(n(n+1)) = 1-1/(n+1)

To prove this by induction:

1. Check to see if it's okay for n = 1.  1/(1.2) = 1-1/2.  Okay.

2. Assume true for n=k, so that 1/(1.2)+...+1/(k(k+1))=1-1/(k+1), and 
   try to prove that it's true for n = k+1, that is,

   1/(1.2)+1/(2.3)+...+1/((k+1)(k+2)) = 1-1/(k+2).

You can do this by just adding 1/((k+1)(k+2)) to both sides of the 
assumed case.                       

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
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