identify the function represented by the following power series. (sum_{k = 1}^{infty}(-1)^{k}\frac{kx^{k +…

Explanation:

Step1: Recall power - series form

Let's start with the geometric - series formula $\sum_{k = 0}^{\infty}r^{k}=\frac{1}{1 - r}$, for $|r|\lt1$. Differentiate both sides with respect to $r$: $\sum_{k = 1}^{\infty}kr^{k - 1}=\frac{1}{(1 - r)^{2}}$, then $\sum_{k = 1}^{\infty}kr^{k}=\frac{r}{(1 - r)^{2}}$.

Step2: Rewrite the given series

The given series is $\sum_{k = 1}^{\infty}(-1)^{k}\frac{kx^{k + 1}}{2^{k}}=x^{2}\sum_{k = 1}^{\infty}k\left(-\frac{x}{2}\right)^{k}$.

Step3: Substitute into the derived formula

Using $\sum_{k = 1}^{\infty}kr^{k}=\frac{r}{(1 - r)^{2}}$ with $r=-\frac{x}{2}$, we have $\sum_{k = 1}^{\infty}k\left(-\frac{x}{2}\right)^{k}=\frac{-\frac{x}{2}}{\left(1+\frac{x}{2}\right)^{2}}$.

Step4: Simplify the expression

$x^{2}\sum_{k = 1}^{\infty}k\left(-\frac{x}{2}\right)^{k}=x^{2}\cdot\frac{-\frac{x}{2}}{\left(1+\frac{x}{2}\right)^{2}}=\frac{-x^{3}}{2\left(1+\frac{x}{2}\right)^{2}}=\frac{-2x^{3}}{(2 + x)^{2}}$, for $\left|-\frac{x}{2}\right|\lt1$ (i.e., $|x|\lt2$).

Answer:

$\frac{-2x^{3}}{(x + 2)^{2}}$