find the power series representation for g centered at 0 by differentiating or integrating the power series…

Explanation:

Step1: Recall geometric - series formula

The geometric - series formula is $\frac{1}{1 - t}=\sum_{k = 0}^{\infty}t^{k}$, for $|t|\lt1$. Let $t = 12x$, then $f(x)=\frac{1}{1 - 12x}=\sum_{k = 0}^{\infty}(12x)^{k}=\sum_{k = 0}^{\infty}12^{k}x^{k}$, for $|12x|\lt1$ or $|x|\lt\frac{1}{12}$.

Step2: Differentiate the power - series of $f(x)$

Differentiate $f(x)=\sum_{k = 0}^{\infty}12^{k}x^{k}$ with respect to $x$. Using the power - rule for differentiation $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f^\prime(x)=\sum_{k = 1}^{\infty}k\cdot12^{k}x^{k - 1}$. Since $g(x)=\frac{12}{(1 - 12x)^{2}}$, and we know that $f^\prime(x)=\frac{12}{(1 - 12x)^{2}}$ (by differentiating $f(x)=\frac{1}{1 - 12x}$ using the quotient rule $\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$ where $u = 1$, $v=1 - 12x$, $u^\prime = 0$, $v^\prime=-12$), so $g(x)=\sum_{k = 1}^{\infty}k\cdot12^{k}x^{k - 1}$. We can rewrite it as $g(x)=\sum_{k = 1}^{\infty}k\cdot12^{k}x^{k - 1}=\sum_{k = 1}^{\infty}k\cdot12^{k - 1}\cdot12x^{k - 1}=\sum_{k = 1}^{\infty}k\cdot12^{k}x^{k - 1}$.

Step3: Find the interval of convergence

The radius of convergence of the geometric series $\sum_{k = 0}^{\infty}(12x)^{k}$ is $R=\frac{1}{12}$ (from $|12x|\lt1$). Differentiating a power - series $\sum_{k = 0}^{\infty}a_{k}x^{k}$ with radius of convergence $R$ does not change the radius of convergence. So the interval of convergence for the power - series of $g(x)$ is $|x|\lt\frac{1}{12}$, or $\left(-\frac{1}{12},\frac{1}{12}\right)$.

Answer:

The power series representation for $g$ is $\sum_{k = 1}^{\infty}k\cdot12^{k}x^{k - 1}$, and the interval of convergence is $\left(-\frac{1}{12},\frac{1}{12}\right)$