Answer
$-1-2(x-1)+3(x-1)^{2}+4(x-1)^{3}+(x-1)^{4}$
$R=\infty$
Work Step by Step
Taylor's series of $f$ centered at $a$ is
$f(x)=\dfrac{f^{n}(a)(x-a)^{n}}{(n)!}=f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^{2}}{2}+\frac{f'''(a)(x-a)^{3}}{6}+....$
Given: $f(x)=x^{4}-3x^{2}+1$
$x^{4}-3x^{2}+1=-1-2(x-1)+\frac{6(x-1)^{2}}{2}+\frac{24(x-1)^{3}}{6}+\frac{24(x-1)^{4}}{24}+0$
$=-1-2(x-1)+3(x-1)^{2}+4(x-1)^{3}+(x-1)^{4}$
$R=\infty$
