Question

Use binomial expansion to tfind the Maclaurin series for f(x)=1/(1+x^9)^2 can somebody tel me if this solution is correct? Sum from 0 to inf of (-1)^n * (1+n)^9n

Answer 1

NO

Answer 2

To check your answer, your series should start like this:

\[1/(81 (x + 1)^2) + 8/(81 (x + 1)) + 88/243 + (20 (x + 1))/27 + \\
172/243 (x + 1)^2 - 104/243 (x + 1)^3 - (4612 (x + 1)^4)/2187 \\- (
3410 (x + 1)^5)/2187 + (22082 (x + 1)^6)/6561+ \cdots \]

Answer 3

I presumed that you want the series around x=-1

Answer 4

Doesn't IxI have to <= 1 to use this series

Answer 5

I meant Sum from 0 to inf of (-1)^n * (1+n)x^9n...

Answer 6

Do they want the Series around zero or around -1

Answer 7

If it is around 0 you do the following
\[
\frac 1 { 1+ u}=\sum_{n=0}^\infty (-1)^{n} u^n, \quad |u|<1\\

-\frac 1 { (1+ u)^2}=\sum_{n=0}^\infty (-1)^{n}n u^{n-1}, \quad |u|<1\\

\frac 1 { (1+ u)^2}=\sum_{n=0}^\infty (-1)^{n+1}n u^{n-1}, \quad |u|<1\\
\frac 1 { (1+ x^9)^2}=\sum_{n=0}^\infty (-1)^{n+1}n (x^9)^{n-1}, \quad |x|<1\\
\frac 1 { (1+ x^9)^2}=\sum_{n=0}^\infty (-1)^{n+1}n x^{9n-9}, \quad |x|<1\\
\]