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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.

mathtoo Dec 28, 2018

#1**+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

CPhill Dec 28, 2018