sum of multiples from 3 to 99

• Feb 5th 2007, 03:33 PM
kcsteven
sum of multiples from 3 to 99
If k=1 is the lower limit and n represents the upper limit and in the middle of the sigma we have (a+(k-1)d) to get the sum of multiples from 3 to 99 confuses me a quite a bit. I know we start at 3 but I am just guessing after that.
Thank You,
Keith stevens:confused:
• Feb 5th 2007, 03:43 PM
topsquark
Quote:

Originally Posted by kcsteven
If k=1 is the lower limit and n represents the upper limit and in the middle of the sigma we have (a+(k-1)d) to get the sum of multiples from 3 to 99 confuses me a quite a bit. I know we start at 3 but I am just guessing after that.
Thank You,
Keith stevens:confused:

You want the sum of the series 3, 6, 9, ..., 96, 99 I presume?

The kth term of the series is 3 + 3(k - 1) where k runs from 1 to 33. So this is:
$\sum_{k = 1}^{33}[ 3 + 3(k - 1)] = 3 \sum_{k = 1}^{33}[1 + k - 1]$

$= 3 \sum_{k = 1}^{33}k$

Now, $\sum_{k = 1}^n k = \frac{n(n+1)}{2}$ so your sum is:
$\sum_{k = 1}^{33}[ 3 + 3(k - 1)] = 3 \cdot \frac{33(33 + 1)}{2} = 1683$

-Dan