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MathSofiya Group TitleBest 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

MathSofiya Group TitleBest 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

MathSofiya Group TitleBest 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

MathSofiya Group TitleBest 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

MathSofiya Group TitleBest 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

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

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

MathSofiya Group TitleBest 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

KingGeorge Group TitleBest 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

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
ok Thanks!
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.2
You're welcome.
 one year ago
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