## anonymous one year ago Help with MacLaurin series. please! Find the MacLaurin series for xarctan(2x) attaching picture below

1. anonymous

this is the actual problem.

2. anonymous

@IrishBoy123

3. IrishBoy123

$$\huge = x \ \Sigma_{n=0}^{n=\infty} \frac{(-1)^n \ (2x)^{2n+1}}{2n + 1}$$ $$\huge = \ \Sigma_{n=0}^{n=\infty} \frac{(-1)^n \ (2)^{2n+1}(x)^{2n+1}x}{2n + 1}$$ $$\huge = \ \Sigma_{n=0}^{n=\infty} \frac{(-1)^n \ (2)^{2n+1}(x)^{2n+2}}{2n + 1}$$ $$\huge = \ \Sigma_{n=0}^{n=\infty} \frac{(-1)^n \ (2)^{2n+1}(x^2)^{n+1}}{2n + 1}$$

4. IrishBoy123

i'm just flailing around here, dude

5. anonymous

I'd made it up to the second step on my own. I totally forgot you could add the exponents of X when multiplying that's where I was stuck. thanks!

6. IrishBoy123

mp