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What is the value of $$1+2+3-4+5+6+7-8+9+10+11+12...+97+98+99-100 \ ?$$ Any help is appreciated, thank you!

I added the terms as an AP then subtracted 10 then all the numbers that were missed out, not sure if this is right though.

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This looks like $(1 + 2 + ... + 100) = 5050$, but with a $-4, -8, -100$, which net a $-8, -16, -200$ to the sum. So $5050- 224$ –  The Chaz 2.0 Mar 21 '12 at 19:26
    
the pattern continues –  Daniel Mar 21 '12 at 19:27
    
the minussing does not only occur in the first few examples –  Daniel Mar 21 '12 at 19:27
    
I didn't see that, since you had $+12$ in there... In that case, take $5050$ and subtract two times (why?) the series $(4 + 8 + 12 + ... + 100)$ –  The Chaz 2.0 Mar 21 '12 at 19:29
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3 Answers

And just to show that there's more than one way to skin a cat: Rearrange to get $$ (1+3+5+\cdots+97+99)+\Big[(2-4)+(6-8)+\cdots+(98-100)\Big] = 2500 + 25\cdot(-2) = 2450$$

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(+1) - also, that expression is gross! –  The Chaz 2.0 Mar 21 '12 at 19:35
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If you rearrange the terms, you can write it as:
$1+2+3+5+6+7+9+10+...+98+99-4-8-...-100=1+2+3+4+...+100-2(4+8+12+...+100)= 1+2+...+100-2\cdot 4(1+2+...+25)=\frac{100(100+1)}{2}-2\cdot4\frac{25(25+1)}{2}$.

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There are $25$ groups of four. The first group has "sum" $2$. The second group has sum $8$ more than the first group. The next group (assuming that $+12$ is a typo for $-12$) has sum $8$ more, and so on.

By the so-called Gauss Method, the full sum is $25$ times the average of first sum and last sum.

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