Can anybody help me with sigma notation please?

Question

anonymous · Answer

How would i evaluate that?

anonymous · Answer

u kno 2(r+1)-1 = 2r+1

anonymous · Answer

how do you get 2(r+1) though? From r = 5?

anonymous · Answer

that's a hint :P

anonymous · Answer

I am confused as i am not sure how to manipulate the sigma...

anonymous · Answer

Do you know where i can get examples?

anonymous · Answer

it's just a shorthand representation of sum...........
nyways write 1st two terms of that xpansion

anonymous · Answer

you mean$$\sum_{6}^{n} (1/2r+1) -\sum_{6}^{n}(1/2r-1)?$$

anonymous · Answer

He means exactly that.

anonymous · Answer

i think i must manipulate the second sigma?

anonymous · Answer

You could further simplify this expression of course. For example, 
$$\sum_{6}^{n}(1/2r+1)$$ could be further reduced to $$1/2 \sum_{6}^{n}n + \sum_{6}^{n}1$$

anonymous · Answer

Excuse me, the n in the previous post I mean to be r

anonymous · Answer

nooooooooooooo
wat's the second term of first xpansion nd first term of second xpansion?

anonymous · Answer

no? sorry, i meant to write $$\sum_{6}^{n}(1/(2r-1))$$

anonymous · Answer

Oh disregard my response then. I didn't see the 1 over expression

anonymous · Answer

you mean 1/11, 1/13?

anonymous · Answer

using your hint 2(r+1)-1 = 2r+1, are you referring to |dw:1339333287054:dw|