Answer
$f(x)=1+3x+6x^2+10x^3+15x^4+....$
Work Step by Step
We have $f'(x)=\dfrac{3}{(1-x)^4} \implies f'(0)=1$;
$f''(x)=\dfrac{12}{x^5}\dfrac{3}{(1-x)^4} \implies f''(0)=3\\ f'''(x)=\dfrac{60}{(1-x)^6}\implies f'''(0)=12$ and $f^{4}(1)=60$ and so on.
Therefore, the Taylor's series at $x=1$ is as follows:
$f(x)=f(0)+f'(0) (x)+\dfrac{f''(0)(x)^2 }{2!}+\dfrac{f''(0)(x)^3 }{3!}+\dfrac{f''(4)(x)^4 }{4!}+.....so \space on$
or, $f(x)=1+3x+6x^2+10x^3+15x^4+....$
