Find the Taylor expansion #\color(red)\bb\text(formula)#... for #f(x)=1/x^2# given #a=4#?

Let's determine the pattern for the #nth# derivative evaluated at #4#. We adopt the convention that the #0th# derivative is just the function itself.

#f(x)=x^-2, f(4)=4^-2#

#f'(x)=-2x^-3, f'(4)=-2(4)^-3#

#f''(x)=(-2)(-3)x^-4, f''(4)=(-2)(-3)4^-4#

#f'''(x)=(-2)(-3)(-4)x^-5, f'''(4)=(-2)(-3)(-4)4^-5#

We can see that we start out with a positive term and then alternate between negative and positive terms, so, we have an instance of #(-1)^n.#

The first exponent on the #4# is #-2#, so, since we're starting at #n=0,# we're going to have a #4^-(n+2)=1/4^(n+2).#

Finally, for the factorial, the 0th derivative has a coefficient of #1#, or #1!#, the first derivative has a coefficient of #-2,# or #-(2!),# the second derivative has a coefficient of #(-2)(-3),# or #3!#.

Knowing that the negatives are handled by the #(-1)^n,# we can represent the factorials by #(n+1)!#

Thus,

#f^((n))(4)=(-1)^n((n+1)!)/4^(n+2)#

Now, using the general form of a Taylor series about #a,#

#sum_(n=0)^oof^((n))(a)(x-a)^n/(n!),# we get

#1/x^2=sum_(n=0)^oo(-1)^n((x-4)^n(n+1)!)/(n!4^(n+2))#

We can simplify the factorials, as #(n+1)! = (n+1)(n!)#

#1/x^2=sum_(n=0)^oo(-1)^n((x-4)^n(n+1)cancel(n!))/(cancel(n!)4^(n+2))#

#1/x^2=sum_(n=0)^oo(-1)^n((x-4)^n(n+1))/(4^(n+2))#