# Sum of Arithmetic Series

• Nov 25th 2009, 02:16 PM
thekrown
Sum of Arithmetic Series
I'm having trouble with this question:

Find the sum of arithmetic series 200 terms, k=1 and on the right of sigma it is 3k+4.

So it's something like: above sigma: 200, below sigma k=1 and on the right (3k+4).

I would edit this post if I could find a way to input those nice graphics with actual math symbols but I don't know where to find it on this site, so I hope it is clear.

term 1 is 3*1 + 4 = 7
term 2 is 3*2+4 = 10
term 3 is 13...

The sum of arithmetic series gives me this:

200/2 * [2x7 + (199)3]
= 6110

• Nov 25th 2009, 03:48 PM
Soroban
Hello, thekrown!

You dropped a zero in the last step . . .

Quote:

Find the sum of arithmetic series: . $\sum^{200}_{k=1}(3k+4)$

first term: $a = 7$
common difference: $d = 3$
no. of terms: $n = 200$

The sum of arithmetic series gives me this:

. . $\tfrac{200}{2}[2(7) + (199)3] \;=\; {\color{blue}(100)}(611) \:=\; {\color{blue}61,\!100}$

• Nov 25th 2009, 04:40 PM
thekrown
Nice! So aside from that small error I got it?

How did you reply with the nice image symbols to display my problem?
• Nov 26th 2009, 03:25 AM
HallsofIvy
Quote:

Originally Posted by thekrown
Nice! So aside from that small error I got it?

How did you reply with the nice image symbols to display my problem?

He is using "LaTex" codes. If you move your pointer over the formula it will turn into a little hand. Clicking on the formula then shows the code used.