The Taylor series for #f(x)# centered at #a# is:

#f(x) = f(a)+f'(a)(x-a)+ (f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+ * * * #

For this question: #f(x) = 1/(x+1)# and #a=2#

#f(x) = 1/(x+1)# #color(white)"sssssssssssss"# so #f(a)=f(2)=1/3#

#f'(x) = -1/(x+1)^2# #color(white)"ssss-s"# so #f'(a)= -1/3^2#

#f''(x)=2/(x+1)^3# #color(white)"sssss-ss"# so #f''(a)= 2/3^3#

#f'''(x)=-6/(x+1)^4# #color(white)"ssss"# so #f'''(a)= -6/3^4#

The general #n-1^(th)# derivative will have

#f^(n-1)(x)=(-1)^(n-1)((n-1)!)/(x+1)^n#

so #f^(n-1)(a)= (-1)^(n-1)((n-1)!)/3^n#

Notice that in each term of the Taylor expansion for this function, I get: #(n-1)!# in both the numerator (from the derivative) and the denominator (from the formula for the expansion). Obviously, these cancel, leaving:

#1/3 - 1/3^2(x-2)+1/3^3(x-2)^2-1/3^4(x-2)^3# for the first four terms, and

#(-1)^(n-1) 1/3^n (x-2)^(n-1)# for the #n^(th)# term.