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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is the actual problem.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\(\huge = x \ \Sigma_{n=0}^{n=\infty} \frac{(1)^n \ (2x)^{2n+1}}{2n + 1}\) \(\huge = \ \Sigma_{n=0}^{n=\infty} \frac{(1)^n \ (2)^{2n+1}(x)^{2n+1}x}{2n + 1}\) \(\huge = \ \Sigma_{n=0}^{n=\infty} \frac{(1)^n \ (2)^{2n+1}(x)^{2n+2}}{2n + 1}\) \(\huge = \ \Sigma_{n=0}^{n=\infty} \frac{(1)^n \ (2)^{2n+1}(x^2)^{n+1}}{2n + 1}\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0i'm just flailing around here, dude

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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!
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