(a) Use differentiation to find a power series representation forf(x) =1(6 + x)2.f(x) =∞ leftparen1.gif(−1)n(n+1)xn6n+2​ rightparen1.gifsum.gifn = 0What is the radius of convergence, R?R = (b) Use part (a) to find a power series forf(x) =1(6 + x)3.f(x) =∞ leftparen1.gif(−1)n(n+3)(n+1)xn6n+5​ rightparen1.gifsum.gifn = 0What is the radius of convergence, R?R = (c) Use part (b) to find a power series forf(x) =x2(6 + x)3.f(x) =∞ leftparen1.gif(−1)n(n+2)(n+1)xn+26n+3​ rightparen1.gifsum.gifn = 2What is the radius of convergence, R?R =

(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].