# Math Help - find sum

1. ## find sum

nb

2. 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

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