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Last edited by oceanmd; Dec 13th 2007 at 04:31 AM.
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Originally Posted by oceanmd $\displaystyle \sum_{n = 1}^{100} \ 3n-11$ Solution: a (first)= -8; a(100th) =289; Sum=14050 Is this correct? Thank you that is correct. did you use the formula for the sum of the first n terms of an arithmetic series here? i suppose it can be done that way, but that's not what i did
Originally Posted by oceanmd I used the formula S= n/2(a(first)+an(n-th term) This is Sum of Finite Arithmatic Series. What did you do? i used the fact that $\displaystyle \sum_{i = 1}^{n}i = \frac {n(n + 1)}2$ so, $\displaystyle \sum_{n = 1}^{100}3n - 11 = 3 \sum_{n = 1}^{100}n - \sum_{n = 1}^{100}11 = 3 \cdot \frac {100(101)}2 - 11(100) = \mbox{you know}$
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