find sum

**oceanmd** · December 6th 2007, 09:41 PM

nb

**Jhevon** · December 6th 2007, 10:54 PM

Originally Posted by oceanmd

$\sum_{n = 1}^{100} \ 3n-11$

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

**Jhevon** · December 6th 2007, 11:11 PM

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 $\sum_{i = 1}^{n}i = \frac {n(n + 1)}2$

so, $\sum_{n = 1}^{100}3n - 11 = 3 \sum_{n = 1}^{100}n - \sum_{n = 1}^{100}11 = 3 \cdot \frac {100(101)}2 - 11(100) = \mbox{you know}$

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