Arithmetic Series 2k: Sum of (-1)^n

  • Thread starter Thread starter resresd
  • Start date Start date
  • Tags Tags
    Series
AI Thread Summary
The discussion focuses on the sum of the series Σ (-1)^n from n=1 to 2k, where 2k indicates an even number of terms. Participants confirm that the sum is indeed 0, as the series alternates between -1 and 1. It is clarified that k is treated as a constant, and the even nature of 2k ensures that all terms cancel out. Suggestions for writing out the series include using words to describe the pattern of terms. The final conclusion is that the sum of the series is 0.
resresd
Messages
10
Reaction score
0
2k
Σ (-1)n
N=1
i am meant to write it out in full and i know that the answer is 0 which i think i understand as it will be -1+1-1+1 etc, but i don't know what to do about the k, it doesn't say that it is a constant, is it safe to assume it is anyway? and what do i do if it is?
thanks
resresd
 
Physics news on Phys.org
(-1)^n, right? Yes, k is a constant. 2k is the number of terms in -1+1-1+1... What's the sum? Hint: 2k is even.
 
yes sorry, i wrote it out in word and copy pasted it, the superscript obviously didn't show up. if 2k showa that the number of terms will be even then with the pattern of -1+1 the answere will be 0...how do i write it out though? is it just -1+1 with a recurring dot? can i do that?
 
The sum is 0, yes. Just write it using words, like '(-1+1)+(-1+1)+... k times =0+0+0+... k times=0'. Something like that.
 
Thanks
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top