BC help on audit exam!

nonexplosive
Sun, 01 May 2011 14:36:02 +0000

This is how I think through problems like 86:

Generally, the 30th degree term of a Taylor polynomial centered at x = 3 would be given by:
[f^(30)(3) * (x-3)^30] / 30!

Looking at the general term of this particular polynomial, the exponent of the (x-3) is 2n. So to get (x-3) ^ 30, we would use n = 15 and get:
[(x-3)^30] / 15!

Set this equal to the previous definition and we get:
[f^(30)(3) * (x-3)^30] / 30! = [(x-3)^30] / 15!

Cancel the (x-3)^30 and multiply both sides by 30! for your final answer:
30! / 15!

Hope that helps!