Question

Help with MacLaurin series. please!
Find the MacLaurin series for xarctan(2x)
attaching picture below

Answer 1

this is the actual problem.

Answer 2

@IrishBoy123

Answer 3

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

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

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

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

Answer 4

i'm just flailing around here, dude

Answer 5

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!

Answer 6

mp