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February 23rd, 2017, 08:16 PM   #11
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Quote:
Originally Posted by romsek View Post
$A + (A+1) + (A+2) \dots + (B-1) + B = $

$\displaystyle{\sum_{k=A}^B}~k = (B-A+1)\left(\dfrac{A+B}{2}\right)$
This is what I'm doing, and it seems to give me a close value to the actual answer, but not the right one, am I doing something wrong?
https://gyazo.com/5e568db79ea159ce7b1befbfd0fbfb94
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February 23rd, 2017, 08:23 PM   #12
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Quote:
Originally Posted by romsek View Post
$A + (A+1) + (A+2) \dots + (B-1) + B = $

$\displaystyle{\sum_{k=A}^B}~k = (B-A+1)\left(\dfrac{A+B}{2}\right)$
Nevermind, I'm just stupid It worked
+rep ! You're a genius.
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February 24th, 2017, 07:28 AM   #13
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Originally Posted by emilo0212 View Post
Yes of course, here you have it . . .
Your example is just a summation of the first 10 natural numbers, and doesn't use its A column.
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