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MathSofiyaBest ResponseYou've already chosen the best response.1
I'll post the remainder of the question in just a second...bear with me
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
From the MacLaurin Table: \[tan^{1}(x)=\sum_{n=0}^{\infty}(1)^n\frac{x^{2n+1}}{2n+1}=x\frac{x^3}{3}+\frac{x^5}{5}\frac{x^7}{7}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
This is what I've come up with so far: \[x^2\sum_{n=0}^{\infty}(1)^n\frac{(x^3)^{2n+1}}{2n+1}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Hmmmm....could it be: \[x^2\sum_{n=0}^{\infty}(1)^n\frac{(x^3)^{2n+3}}{2n+1}\]?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I meant: \[\sum_{n=0}^{\infty}(1)^n\frac{(x^3)^{2n+3}}{2n+1}\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
I think both the first and third solutions you have are correct.
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
I am inclined to agree
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Yes! So this would be correct? \[\sum_{n=0}^{\infty}(1)^n\frac{(x^3)^{2n+3}}{2n+1}\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
Looks good to me. If you wanted to, you could also say\[(x^3)^{2n+3}=x^{6n+9}\]
 one year ago
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