(a) Wild guess:
[tex]f(x)=\dfrac1{(6+x)^2[/tex]
Recall the power series
[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]
With some manipulation, we can write
[tex]\displaystyle\frac1{6+x}=\frac16\frac1{1-\left(-\frac x6\right)}=\frac16\sum_{n=0}^\infty\left(-\frac x6\right)^n=\sum_{n=0}^\infty\frac{(-x)^n}{6^{n+1}}[/tex]
Take the derivative and we get
[tex]\displaystyle-\frac1{(6+x)^2}=-\sum_{n=0}^\infty\frac{n(-x)^{n-1}}{6^{n+1}}[/tex]
[tex]\displaystyle=-\sum_{n=1}^\infty\frac{n(-x)^{n-1}}{6^{n+1}}[/tex]
[tex]\displaystyle=-\sum_{n=0}^\infty\frac{(n+1)(-x)^n}{6^{n+2}}[/tex]
so we have
[tex]\displaystyle\frac1{(6+x)^2}=\sum_{n=0}^\infty\frac{(n+1)(-x)^n}{6^{n+2}}[/tex]
By the ratio test, this series converges if
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+2)(-x)^{n+1}}{6^{n+3}}}{\frac{(n+1)(-x)^n}{6^{n+2}}}\right|=\left|\frac x6\right|\lim_{n\to\infty}\frac{n+2}{n+1}=\left|\frac x6\right|<1[/tex]
or [tex]|x|<6[/tex], so that the radius of convergence is [tex]R=6[/tex].
(b). If we take the second derivative, we get
[tex]\displaystyle\frac2{(6+x)^3}=\sum_{n=0}^\infty\frac{n(n+1)(-x)^{n-1}}{6^{n+2}}[/tex]
[tex]\displaystyle=\sum_{n=1}^\infty\frac{n(n+1)(-x)^{n-1}}{6^{n+2}}[/tex]
[tex]\displaystyle=\sum_{n=0}^\infty\frac{(n+1)(n+2)(-x)^n}{6^{n+3}}[/tex]
[tex]\displaystyle\frac1{(6+x)^3}=\frac12\sum_{n=0}^\infty\frac{(n+1)(n+2)(-x)^n}{6^{n+3}}[/tex]
Apply the ratio test again and we get [tex]R=6[/tex].
(c) Multiply the previous series by [tex]x^2[/tex] and we get
[tex]\displaystyle\frac{x^2}{(6+x)^3}=\frac12\sum_{n=0}^\infty\frac{(n+1)(n+2)(-x)^nx^2}{6^{n+3}}[/tex]
[tex]\displaystyle=\frac12\sum_{n=0}^\infty\frac{(n+1)(n+2)(-1)^nx^{n+2}}{6^{n+3}}[/tex]
The ratio test yet again tells us [tex]R=6[/tex].
