attempt 2: the answer is incorrect. 8 attempts remaining. given that ∑(n = 0)^∞ x^n = 1/(1 - x), for |x| <…

Explanation:

Step1: Start with the given series

We know that $\sum_{n = 0}^{\infty}x^{n}=\frac{1}{1 - x}$, for $|x|<1$. Replace $x$ with $x^{2}$, we get $\sum_{n = 0}^{\infty}(x^{2})^{n}=\frac{1}{1 - x^{2}}$, for $|x^{2}|<1$ (i.e., $|x|<1$). So $\sum_{n = 0}^{\infty}x^{2n}=\frac{1}{1 - x^{2}}$.

Step2: Differentiate both - sides

Differentiate $\sum_{n = 0}^{\infty}x^{2n}$ and $\frac{1}{1 - x^{2}}$ with respect to $x$. The derivative of $\sum_{n = 0}^{\infty}x^{2n}$ term - by - term is $\sum_{n = 1}^{\infty}2nx^{2n - 1}$ (using the power rule $\frac{d}{dx}(x^{m})=mx^{m - 1}$). The derivative of $\frac{1}{1 - x^{2}}=(1 - x^{2})^{-1}$. Using the chain rule, if $u = 1 - x^{2}$, then $\frac{d}{dx}(1 - x^{2})^{-1}=(-1)(1 - x^{2})^{-2}(-2x)=\frac{2x}{(1 - x^{2})^{2}}$.

Step3: Re - index if necessary

The power series for $f(x)=\frac{2x}{(1 - x^{2})^{2}}$ is $\sum_{n = 1}^{\infty}2nx^{2n - 1}$. Since the index starts from $n = 1$, we can rewrite it as $\sum_{n = 0}^{\infty}2(n + 1)x^{2(n+1)-1}=\sum_{n = 0}^{\infty}2(n + 1)x^{2n+1}$.

Answer:

$\sum_{n = 1}^{\infty}2nx^{2n - 1}$