Representing a function as a power series and finding the coefficients of the terms? My prblem asks me to find the coefficients of the first five terms of the power series representation of the function:
$$f(x)=\frac{10}{17+x}$$
$$f(x)=10*\frac{\frac{1}{17}}{1-\frac{-x}{17}}$$
$$\sum_{n=0}^{\infty}10*\frac{1}{17}*(\frac{-x}{17})^n$$
$$\sum_{n=0}^{\infty}\frac{10}{17}*\frac{(-1)^nx^n}{17^n}$$
First five terms are:
$$\frac{10}{17} + \frac{10}{17}*\frac{-x}{17} + \frac{10}{17}*\frac{x^2}{17^2} + \frac{10}{17}*\frac{-x^3}{17^3} + \frac{10}{17}*\frac{x^4}{17^4}$$
I entered $0$ for the first coefficient. Marked me correct.
I put $\frac{10}{17}$ for the second coefficient. Marked me correct.
I put $\frac{10}{17}$ for the third coefficient. Marked me wrong.
I'm not sure what I'm doing wrong. Looking at the series again, the coefficient should be $-\frac{10}{17}$ for the second coefficient, shouldn't it?
I tried a bunch of stuff for the second coefficient too. $0$, $-1$, $1$, $-10$, $10$, $\frac{10}{17}$, $-\frac{10}{17}$, $\frac{100}{17^2}$, $-\frac{100}{17^2}$, $\frac{10}{17^3}$, $-\frac{10}{17^3}$ (those last few are from different ways of writing the power series I tried). Nothing is working. Where am I going wrong?
Edit: to clarify a bit, the question wants me to find the coefficients of $x^0$, $x$, $x^2$, $x^3$ and $x^4$ in the series.
 A: The coefficients of the terms of a Maclaurin power series for a function $f(x)$ are numbers $a_0,a_1,a_2,\dots$ such that $$f(x)=\sum_{n=0}^\infty a_nx^n$$ for all $x$ within the interval of convergence of the series. You got very close to finding them, but unfortunately, didn't quite get all the way there, because you didn't get the powers of $x$ separated from the rest. Since
\begin{eqnarray}f(x) &=& \sum_{n=0}^\infty\frac{10}{17}\cdot\frac{(-1)^nx^n}{17^n}\\ &=& \sum_{n=0}^\infty\frac{10\cdot(-1)^nx^n}{17\cdot17^n}\\ &=& \sum_{n=0}^\infty\frac{10\cdot(-1)^nx^n}{17^{n+1}}\\ &=& \sum_{n=0}^\infty\frac{10\cdot(-1)^n}{17^{n+1}}x^n,\end{eqnarray} then the coefficients in general have the form $\frac{10\cdot(-1)^n}{17^{n+1}}$ for all $n.$ Hence, the coefficients you're looking for are $$\frac{10}{17},-\frac{10}{17^2},\frac{10}{17^3},-\frac{10}{17^4},\frac{10}{17^5}.$$ I have no idea why $0$ would've been accepted as the first coefficient.

Edit: Thanks to your comment below, I see that the actual function in question is $$g(x)=\frac{10x}{17+x}.$$ Letting $f(x)=\frac{10}{17+x}$ be the initially posted function, we see that $g(x)=x\cdot f(x),$ and so $$\begin{eqnarray}g(x) &=& x\cdot\sum_{n=0}^\infty\frac{10\cdot(-1)^n}{17^{n+1}}x^n\\ &=& \sum_{n=0}^\infty\frac{10\cdot(-1)^n}{17^{n+1}}x^{n+1}\\ &=& \sum_{n=1}^\infty\frac{10\cdot(-1)^{n-1}}{17^n}x^n\\ &=& 0x^0+\sum_{n=1}^\infty\frac{10\cdot(-1)^{n-1}}{17^n}x^n,\end{eqnarray}$$ whence the first five coefficients are $$0,\frac{10}{17},-\frac{10}{17^2},\frac{10}{17^3},-\frac{10}{17^4}.$$
A: You allready found $$f(x)=\frac{10}{17} + \frac{10}{17}\frac{-x}{17} + \frac{10}{17}\frac{x^2}{17^2} + \frac{10}{17}*\frac{-x^3}{17^3} + \frac{10}{17}\frac{x^4}{17^4}+...$$
$$=\frac{10}{17} - \frac{10}{17^2}x + \frac{10}{17^3}x^2 - \frac{10}{17^4}x^3 + \frac{10}{17^5}x^5 \mp ...$$
The general coefficient $a_n$ is given by
$$a_n=(-1)^n\dfrac{10}{17^{n+1}}.$$
