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avatar+809 

Compute the value of \(N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2,\) where the additions and subtractions alternate in pairs.

 Dec 28, 2018
 #1
avatar+98102 
+2

Note that we can rearrange this as

 

100^2 - 1^2  +  99^2 - 2^2  + 3^2 - 98^2 + 4^2 - 97^2 + .....

 

(100+ 1) (100 - 1) + ( 99 + 2) (99 - 2) + (3 + 98) (3 -  98) + (4 + 97)(4 - 97) +....

 

(101) (99 + 97 - 95 - 93.....)  =

 

(101) (99 - 95 + 97 -  93..... )   =

 

(101) ( 4 + 4.....)   

 

We paired 4 terms to get one 4

 

We paired 8 terms to get two 4's

 

So......pairing 100 terms will yield twenty-five 4's

 

So.....the sum is  (101)(25 * 4)  =  101 * 100 = 10100

 

 

cool cool cool

 Dec 28, 2018
edited by CPhill  Dec 28, 2018
edited by CPhill  Dec 28, 2018
edited by CPhill  Dec 28, 2018
edited by CPhill  Dec 28, 2018
 #2
avatar+809 
+1

Okay, thanks so much, CPhill!

mathtoo  Dec 29, 2018
 #3
avatar
0

This forms an arithmetic series as follows:

First term =788, Number of terms =25, Common difference =-32

SUM = N/2 * [2*F + D*(N - 1)]

SUM = 25/2 *[1,576 + (-32 * 25 - 1]

SUM = 12.5 *[1,576 + (-32 *24]

SUM = 12.5 *[1,576 - 768]

SUM = 12.5 *[808]

SUM =10,100

 Dec 29, 2018

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