Difficult Taylor Series approximation?

Question

I need to find an approximation for (2x-3)/(x+4) centered about 2.

So far, I have

f(c)=1/6

f'(c)=11/6^2

f''(c)=-22/6^3

f'''(c)=66/6^4

etc.

So my polynomial is:

1/6 + 11(x-2)/6^2 - 22(x-2)^2 /6^3 + 66(x-2)^3 / 6^4 etc.

I am having difficulty finding the Kth term.

I toyed around with (x-2)^k/6^(k+1) but can't figure out how to get the alternating series portion started because of the two positive integers for k=0 and k=1. Also, I'm not sure how to work in the 11 term that begins at k=1.

I would appreciate help with this.

Answer 1

so write (2x-3) / ( x + 4)  as ( 2[x-2] +1) / ( 6 +[x-2] )  = {[1/3][x-2] / ( 1 + [x-2] / 6 ) } + {[1/ 6] / ( 1 + [x-2] / 6) }...you do know that 1 / [1+t] = Σ (-1)^n t^n...so the 1st part is [1/3][x-2] Σ (-1)^n [x-2]^n / 6^n , for |x-2| < 6...this should help you..you finish