Question

Unsure....
\[f(x)=x^2tan^{-1}(x^3)\]

Answer 1

I'll post the remainder of the question in just a second...bear with me

Answer 2

From the MacLaurin Table:
\[tan^{-1}(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\]

Answer 3

This is what I've come up with so far:
\[x^2\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+1}}{2n+1}\]

Answer 4

Hmmmm....could it be:
\[x^2\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}\]?

Answer 5

I meant:
\[\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}\]

Answer 6

I think both the first and third solutions you have are correct.

Answer 7

I am inclined to agree

Answer 8

Yes!
So this would be correct?
\[\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}\]

Answer 9

Looks good to me. If you wanted to, you could also say\[(x^3)^{2n+3}=x^{6n+9}\]

Answer 10

ok Thanks!

Answer 11

You're welcome.