Two Ways to Find a Formula

Date: 04/14/98 at 17:24:19
From: Mon
Subject: Calculus, differentiating, geometric series

For any real number x, such that x does not equal 1, I am trying to 
prove that:

     n          x-(n+1)x^(n+1) + nx^(n+2)
   Sigma rx^r = -------------------------
    r=1                   (1-x)^2

by using calculus, differentiating (with respect to x), and the 
formula for the sum of the geometric series:

     n
   Sigma x^r   
    r=0

Date: 04/15/98 at 08:02:27
From: Doctor Anthony
Subject: Re: Calculus, differentiating, geometric series

The series required is:

   x + 2x^2 + 3x^3 + 4x^4 + ... + nx^n

You can sum this in two ways. The non-calculus way is as follows:

        S =  x + 2x^2 + 3x^3 + ... + nx^n
       xS =       x^2 + 2x^3 + ... + (n-1)x^n + nx^(n+1)
   ----------------------------------------------------- subtracting
   (1-x)S =  x +  x^2 +  x^3 + ... + x^n      - nx^(n+1)

   (1-x)S = x(1-x^n)/(1-x)  -  nx^(n+1)
 
            x(1-x^n)      nx^(n+1)
        S = --------  -  ----------
            (1-x)^2        (1-x)

            x(1-x^n) - n(1-x)x^(n+1)
        S = ------------------------
                   (1-x)^2

            x - x^(n+1) - nx^(n+1) + nx^(n+2)
        S = ---------------------------------
                     (1-x)^2

             x - (n+1)x^(n+1) + nx^(n+2)
        S = ----------------------------
                     (1-x)^2

CALCULUS METHOD

Consider the series:

   1 + x + x^2 + ......+ x^n

differentiating:

   1 + 2x + 3x^2 + 4x^3 + .. + nx^(n-1)

multiply by x:

   x + 2x^2 + 3x^3 + ........+ nx^n

The sum of the series 1 + x + x^2 + .... + x^n is given by:

   1-x^(n+1)
   ---------
     1-x
                                 
and differentiating this we get:

   (1-x)(-(n+1)x^n) - (1-x^(n+1))(-1)
   ----------------------------------
              (1-x)^2

     -(n+1)x^n + (n+1)x^(n+1) + 1 - x^(n+1)
   = --------------------------------------
                 (1-x)^2  

     1 - (n+1)x^n + nx^(n+1)
   = -----------------------
           (1-x)^2

and finally to get the sum we require we must multiply this result
by x.

The required sum is: 

   x - (n+1)x^(n+1) + nx^(n+2)
   ---------------------------
            (1-x)^2

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