+0  
 
0
202
2
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(204+202+200+......+6+4+2)-(203+201+199+......+5+3+1) =?

Guest Oct 26, 2014

Best Answer 

 #1
avatar+26402 
+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
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2+0 Answers

 #1
avatar+26402 
+10
Best Answer

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
avatar+81023 
+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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