+0

# Values

0
130
3
+814

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
+100564
+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

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
+814
+1

Okay, thanks so much, CPhill!

mathtoo  Dec 29, 2018
#3
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