Expert: Paul Klarreich - 10/29/2008

Question
1) solve (1/3)+(1/15)+(1/35)+...(1/(2(n-1)(2n+1))=(1/(2n+1)).
2) solve 1+2+2^2+...+2^(n-1)=2^(n)-1

Answer
Questioner:   tofunmi
Category:  Advanced Math
Private:  No
 
Subject:  mathematical induction
Question:  1) solve (1/3)+(1/15)+(1/35)+...(1/(2(n-1)(2n+1))=(1/(2n+1)).
2) solve 1+2+2^2+...+2^(n-1)=2^(n)-1
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Hi,Tofunmi,

Here is the scheme for doing M.I.  (BTW, don't ever use those letters if you are talking to your doctor.)

Phase 1. WRITE the theorem for  n = the 'base case', usually, but not always, n = 1.
Check that it is correct.

Phase 2a. WRITE the theorem for  n = k.  This is the ASSUMPTION.

Phase 2b.WRITE the theorem for  n = (k+1).  This is the STATEMENT TO PROVE.  

(and be sure to use parentheses around the k+1.)
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NOW a couple of tricks for handling a proof involving a SUMMATION, like yours:

The SUM for  n=1  will almost certainly contain only one term, with no dots. ...

The SUM for  n=k+1  will almost certainly CONTAIN the sum for n = k, and just have one extra term.  You will replace the  sum for n=k with the 'right side' of your ASSUMPTION.
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I'll get you started and you will finish up.

WARNING: THIS DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL DIFFICULT TO VIEW ON CERTAIN COMPUTING SYSTEMS.  VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

BTW, you 'PROVE' these statements, you don't 'solve' them.

Oh, yes.  One little thing -- it helps if you write the theorem correctly.

You wrote:
Theorem 1:  (1/3)+(1/15)+(1/35)+...(1/(2(n-1)(2n+1))=(1/(2n+1)).

That is wrong.  The correct statement is:
Theorem 1:  (1/3)+(1/15)+(1/35)+...(1/((2n-1)(2n+1))= n/(2n+1).

(it looks a lot nicer and cleaner like this:)

 1       1       1                1            n
----- + ----- + ----- + ... + ------------ = ---------
1*3     3*5     5*7          (2n-1)(2n+1)    2n + 1

OK, we can start:

Phase 1:  n = 1:  
 1          1
-----  = ---------
1*3      2(1)+ 1

1/3 = 1/3,  check.

Phase 2a:

1     1     1               1             k
--- + --- + --- + ... + ------------ = ---------  Assumed
1*3   3*5   5*7         (2k-1)(2k+1)    2k + 1

1     1     1               1                  1                k+1
--- + --- + --- + ... + ------------ + -------------------- = -----------  To be proved
1*3   3*5   5*7         (2k-1)(2k+1)   (2(k+1)-1)(2(k+1)+1)   2(k+1) + 1

use the assumption:
<============ all this part=========>
1     1     1               1                  1                k+1
--- + --- + --- + ... + ------------ + -------------------- = -----------  To be proved
1*3   3*5   5*7         (2k-1)(2k+1)   (2(k+1)-1)(2(k+1)+1)   2(k+1) + 1

<===becomes this by assumption =====>
                 k                             1                k+1
           ------------ +             ------------------- = -----------  To be proved
              2k + 1                   (2k+2-1)(2k+2)+1)     2(k+1) + 1


                 k                             1                k+1
           ------------ +             ------------------- = -----------  To be proved
              2k + 1                    (2k+1)(2k+3)           2k+3

I leave the rest to you -- combine the fractions on the left and simplify.  If you don't slip up, it comes out to the same as the right side.


Theorem 2) prove  1 + 2 + 2^2 + ... + 2^(n-1)= 2^(n)-1

I think you can try the same scheme on this.  If you get stuck, send me what you did and I'll try to find where you went wrong.