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Power Sum problem

Find the sum of 1^2 - 2^2 + 3^2 - 4^2 + \ldots + (-1)^{m-1} \cdot m^2

Solution:

Let the sum be S_m . If m is even, then :

S_m = 1^2 - 2^2 + 3^2 -4^2 + \ldots + (m-1)^2 - m^2 = - [ 3 + 7 + \ldots + (2m-1) ] = -m(m+1)/2

If m is odd, then :

S_m = S_{m-1} + m = - \frac{ (m-1)m }{2} + m^2 = m(m+1)/2

S_m = (-1)^{m-1} m(m+1)/2

 

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Categories: Power Sums Tags: ,
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  1. November 27, 2011 at 18:45
  2. December 20, 2011 at 11:39

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