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

Explanation:

Step1: Rewrite the power - series

Let (S=\sum_{k = 1}^{\infty}(-1)^{k}\frac{kx^{k + 1}}{3^{k}}=x^{2}\sum_{k = 1}^{\infty}k(-\frac{x}{3})^{k}).

Step2: Recall the formula for (\sum_{k = 1}^{\infty}kx^{k})

We know that (\sum_{k = 0}^{\infty}x^{k}=\frac{1}{1 - x}), for (|x|\lt1). Differentiating both sides with respect to (x), we get (\sum_{k = 1}^{\infty}kx^{k - 1}=\frac{1}{(1 - x)^{2}}), then (\sum_{k = 1}^{\infty}kx^{k}=\frac{x}{(1 - x)^{2}}) for (|x|\lt1).

Step3: Substitute (x=-\frac{x}{3}) into the formula

Let (t =-\frac{x}{3}), then (\sum_{k = 1}^{\infty}k(-\frac{x}{3})^{k}=\frac{-\frac{x}{3}}{(1+\frac{x}{3})^{2}}=\frac{-x}{3(1 +\frac{x}{3})^{2}}=\frac{-x}{3\times\frac{(3 + x)^{2}}{9}}=\frac{-3x}{(x + 3)^{2}}).

Step4: Find the function (f(x))

Since (S=x^{2}\sum_{k = 1}^{\infty}k(-\frac{x}{3})^{k}), then (f(x)=\frac{-3x^{3}}{(x + 3)^{2}}), for (|-\frac{x}{3}|\lt1) (i.e., (|x|\lt3)).

Answer:

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