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on this one i just need to know how to set it up...
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\[\sum_{k=1}^{50}[(k+1)^2-k^2]\]
i'll solve it if i can just get help setting it up...
your first step is to simplify the equation: (k+1)(k+1) - k^2 (multiply it all out) what do you get?
2k+1
great! so now it's a much easier problem. and you'll do it just like the last problem you did.
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the sum of a constant is just that constant right?
\[ \sum_{j=1}^n c = nc \]
so my final answer is.... 2600
Yes
SWEET!!!!
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\[ \sum_{k=1}^{50} (2k+1) = 2 \sum_{k=1}^{50} k + \sum_{k=1}^{50} 1 = 2 \frac{50(50+1)}{2} + 50 = 2550 + 50 = 2600. \]
that is exactly what i did
good
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