Answer
$\frac{1}{3}$, - $\frac{x-2}{9}$, $\frac{(x-2)^{2}}{27}$, -$\frac{(x-2)^{3}}{81}$
Work Step by Step
$f(x)=\frac{1}{1+x}$ ; $f(2)=\frac{1}{3}$
$f'(x)=\frac{-1}{(1+x)^2}$ ; $f'(2)=-\frac{1}{9}$
$f''(x)=\frac{2}{(1+x)^3}$ ; $f''(2)=\frac{2}{27}$
$f'''(x)=\frac{-6}{(1+x)^4}$ ; $f'''(2)=-\frac{2}{27}$
n-th term of Taylor Series is $\frac{f^{n-1}(2)(x-2)^{n-1}}{(n-1)!}$
1st term of Taylor series is $\frac{f(2)(x-2)^{0}}{(0)!} = \frac{1}{3}$
2nd term of Taylor series is $\frac{f'(2)(x-2)^{1}}{(1)!} = -\frac{(x-2)}{9}$
3rd term of Taylor series is $\frac{f''(2)(x-2)^{2}}{(2)!} = \frac{(x-2)^2}{27}$
4th term of Taylor series is $\frac{f'''(2)(x-2)^{3}}{(3)!} = -\frac{(x-2)^3}{81}$
Hence the first four non-zero terms of the Taylor's Series are:
$\frac{1}{3}$, - $\frac{x-2}{9}$, $\frac{(x-2)^{2}}{27}$, -$\frac{(x-2)^{3}}{81}$
