I guess the function is
[tex]f(x)=\dfrac6{(1-x)^2}[/tex]
Rather than computing derivatives of [tex]f[/tex], recall that for [tex]|x|<1[/tex], we have
[tex]g(x)=\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n[/tex]
Notice that
[tex]g'(x)=\dfrac1{(1-x)^2}[/tex]
so that [tex]f(x)=6g'(x)[/tex]. Then
[tex]f(x)=6\displaystyle\sum_{n=0}^\infty nx^{n-1}=6\sum_{n=1}^\infty nx^{n-1}=6\sum_{n=0}^\infty(n+1)x^n[/tex]
also valid only for [tex]|x|<1[/tex], so that the radius of convergence is 1.
