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Proof of Series Sum


Date: 06/19/97 at 01:16:59
From: piggy
Subject: mathematics

How do you prove:

1x2 + 2x3 + 3x4 + 4x5 + 5x6 + 6x7......+n(n+1) = (nx(n+1) x (n+2))/3?


Date: 06/19/97 at 05:05:52
From: Doctor Anthony
Subject: Re: mathematics

Use SIGMA(1 to n)[r(r+1)]

    SIGMA[r^2 + r]  =  n(n+1)(2n+1)/6  +  n(n+1)/2

                    = [n(n+1)/6][2n+1 + 3]

                    = [n(n+1)/6][2n+4]

                    = [n(n+1)(n+2)]/3


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


Date: 06/28/97 at 12:09:11
From: peggy
Subject: Re: mathematics

Dr. Math,

Thank you for your response.  However, I still do not understand how 
to prove it. Can you explain it in details? (The solution uses SIGMA. 
I don't know what this is.)


Date: 06/28/97 at 18:16:15
From: Doctor Anthony
Subject: Re: mathematics

I was using the standard formulae for the sum 1^2 + 2^2 + 3^2 + .. + 
n^2 and for 1 + 2 + 3 + ... + n.  However if you are not familiar with 
these we can prove the result as follows.

Consider n(n+1)(n+2) - (n-1)n(n+1) =  n(n+1)[n+2-n+1] = 3n(n+1)

So we now consider a series of equations as follows:

    n(n+1)(n+2) - (n-1)n(n+1)     =  3n(n+1)
   (n-1)n(n+1)  - (n-2)(n-1)(n)   =  3(n-1)n  
   (n-2)(n-1)n  - (n-3)(n-2)(n-1) =  3(n-2)(n-1)
    -----------------------------    -----------
    -----------------------------    -----------
    3 x 4 x 5   -  2 x 3 x 4      =  3(3)(4)
    2 x 3 x 4   -  1 x 2 x 3      =  3(2)(3)        Add all the
    1 x 2 x 3   -  0 x 1 x 2      =  3(1)(2)        equations       
   -----------------------------------------------   
    n(n+1)(n+2) - 0               = 3[1x2 + 2x3 + ... + n(n+1)]

Note that when adding all the equations there is a cancelling of terms 
on the lefthand side between adjacent lines. The term -(n-1)n(n+1) in 
the first line cancels with the term +(n-1)n(n+1) in the second line, 
and so on down all the equations.  The only terms that do not cancel 
are the first term in the top equation and the second term in the last 
equation.

             n(n+1)(n+2)
We then get  -----------  = 1x2 + 2x3 + 3x4 + .... + n(n+1)  
                  3

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

    
Associated Topics:
High School Sequences, Series

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