+0

# (204+202+200+......+6+4+2)-(203+201+199+......+5+3+1)

0
201
2

(204+202+200+......+6+4+2)-(203+201+199+......+5+3+1) =?

Guest Oct 26, 2014

#1
+26397
+10

Rewrite as:

(204-203) + (202-201) + ... + (6-5) + (4-3) + (2-1)

1        +       1       + ... +    1    +    1    +   1

There are 102 terms so the sum is 102.

.

Alan  Oct 26, 2014
Sort:

#1
+26397
+10

Rewrite as:

(204-203) + (202-201) + ... + (6-5) + (4-3) + (2-1)

1        +       1       + ... +    1    +    1    +   1

There are 102 terms so the sum is 102.

.

Alan  Oct 26, 2014
#2
+80874
+5

Another way to see this is to note that that the sum of the first n even positive integers is n(n+1)

And the sum of the first n positive odd integers is n^2.

Therefore

n(n+1) - n^2   = n^2 + n - n^2  =  n

And since n = 102 in both cases, then that's the final sum.

CPhill  Oct 26, 2014

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