Why Do Taylor Series Representations of Cosine Use Alternating Powers of -1?

[nhrock3]

when i develop the series of a cosine i have a (-1) member
i wanted to represent the series as a sum
so i need to take only the odd members so the power of -1 is 2k+1 i got
but the solution says that the power of -1 is equal (-1)^{k-1}

is it the same??
why they have such an expression
(they use n istead of k)
http://i45.tinypic.com/6sszue.jpg

how they got the power?

 

[Mark44]

You want the signs of your Taylor's series to alternate, right? (-1)^{k - 1} gives you that sign alternation. If you had (-1)^{2k + 1}, the sign would always be negative, since you have odd powers of -1.

 

[nhrock3]

you are correct
how to get this expression?

 

[Mark44]

What exactly are you asking? Are you asked to find the Taylor's series for cos z at z = 2? There is a standard technique for finding the coefficients of this series.