Answer
$1+3x+6x^2+10x^3+15x^4+....$
Work Step by Step
Differentiate the given function $f(x)$ as follows:
$f'(x)=\dfrac{3}{(1-x)^4} \implies f'(0)=1 \\ f''(x)=\dfrac{12 \times 3}{x^5 (1-x)^4} \implies f''(0)=3\\ f'''(x)=\dfrac{60}{(1-x)^6}\implies f'''(0)=12$
The Taylor series at $x=1$ can be written as:
$f(x)=f(0)+f'(0) (x)+\dfrac{f''(0)}{2!} (x)^2 +\dfrac{f''(0)}{3!} (x)^3 +\dfrac{f''(4) }{4!}(x)^4 +..... \\=1+3x+6x^2+10x^3+15x^4+....$
