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Compute the sum \( 101^2 - 97^2 + 93^2 - 89^2 + \cdots + 5^2 - 1^2\) by pairing the terms and applying the difference of squares.

 Jul 9, 2020
edited by mathmathj28  Jul 9, 2020
edited by mathmathj28  Jul 9, 2020
edited by mathmathj28  Jul 9, 2020
 #1
avatar+21958 
+3

1012 - 972  =  (101 - 97)(101 + 97)  =  4·198

  932 - 892  =  (93 - 89)(93 + 89)      =  4·182

  852 - 812  =  (85 - 81()(85 + 81)     =  4·166

   ...

      52 - 12  =  (5 -1)(5 + 1)               =  4·6 

 

Sum  =  4·198  +  4·182  +  4·166  +  ...  +  4·6  =  4·[ 198 + 182 + 166 + ... + 6 ]

 

198 + 182 + 166 + ... + 6  is an arithmetic series whose first term = 198, last term = 6, common difference =  -16

 

The formula for the sum is:  Sum  =  n · (t1 + tn) / 2

 

We'll need to find the number of terms; we can do this by using this formula:  tn  =  t1 + (n - 1)·d

   --->   6  =  198 + (n - 1)·-16   --->   -192  =  (n - 1)·-16   --->   12  =  n - 1   --->   n  =  13

 

Sum  =  n · (t1 + tn) / 2   --->   Sum  =  13 · (198 + 6) / 2    --->   1326

 

So, 4·[ 198 + 182 + 166 + ... + 6 ]  =  4·1326  =  5304

 Jul 9, 2020
 #2
avatar+313 
+2

Thank you so much for the explanation!!!

mathmathj28  Jul 9, 2020

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