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find the Maclaurin series of $$\frac{1}{1+4x^2}$$ and $$\frac{1}{1+9x^2}$$ using $$\arctan x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$$
Firstly I derived the formula they gave us, then substituted $x \rightarrow 9x$. Is this the right approach?
As you know that: $$ \arctan(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \quad\Rightarrow\quad \arctan (2x) = \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{2n+1} $$ Also, you have: $$ \frac{d}{dx}\left[\arctan(x)\right] = \frac{1}{1+x^2} \quad\Rightarrow\quad \frac{d}{dx}\left[\arctan(2x)\right] = \frac{2}{1+4x^2} $$ Thus, $$ \begin{align} \color{red}{\frac{1}{1+4x^2}} &= \frac{1}{2}\,\frac{d}{dx} \left[\color{white}{\frac{}{}}\arctan(2x)\color{white}{\frac{}{}}\right] = \frac{1}{2}\,\frac{d}{dx} \left[\,\sum_{n=0}^{\infty}(-1)^n\,\frac{(2x)^{2n+1}}{2n+1}\,\right] \\[2mm] &= \frac{1}{2}\,\sum_{n=0}^{\infty}(-1)^n\,\frac{\left(\,(2x)^{2n+1}\,\right)'}{2n+1} \\[2mm] &= \frac{1}{2}\,\sum_{n=0}^{\infty}(-1)^n\,\frac{{2\,(2n+1)}\,(2x)^{2n}}{2n+1} \\[2mm] &= \color{red}{\sum_{n=0}^{\infty}(-1)^n\,(2x)^{2n}} \end{align} $$ And with similar analysis, you should get: $$ \frac{1}{1+9x^2}=\sum_{n=0}^{\infty}(-1)^n\,(3x)^{2n} $$
