equating coefficients in algebraic expansion If $\displaystyle \bigg(\frac{1+x}{1-x}\bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots +\infty,$ then value of 
$(1)\; \displaystyle \frac{3b_{3}-b_{1}}{b_{2}}$
$(2)\; \displaystyle \frac{2b_{4}-b_{2}}{b_{3}}$
$(3)\; \displaystyle \frac{3b_{6}-2b_{4}}{b_{5}}$
$(4)\; \displaystyle \frac{5b_{10}-4b_{8}}{b_{9}}$
Answers given in $n,2n,3n,4n$ formats
What I tried: 
$\displaystyle (1+x)^n(1-x)^{-n}=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots \cdots \infty$
$\displaystyle \bigg[1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}\cdots \bigg]\bigg[1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+\cdots \bigg]=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots $
$\displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$
How do I find other coefficients?  Help me, please.
 A: You made a mistake. It is not $\displaystyle b_{1}=n,b_{2}=n,b_{3}=4n^2$ , but
$b_{1}=2n$
$b_{2}=2n^2$
$b_{3}=\frac23(n+2n^3)$
To go up to $b_{10}$, the series has to be expended up to $x^{10}$. 
By brut force :
$b_{4}=\frac23(2n^2+n^4)$
$b_{5}=\frac{2}{15}(3n+10n^3+2n^5)$
$b_{6}=\frac{2}{45}(23n^2+20n^4+2n^6)$
$b_{7}=\frac{2}{315}(45n+196n^3+70n^5+4n^7)$
$b_{8}=\frac{2}{315}(132n^2+154n^4+28n^6+n^8)$
$b_{9}=\frac{2}{2835}(315n+1636n^3+798n^5+84n^7+2n^8)$
$b_{10}=\frac{2}{14175}(5167n^2+7180n^4+1806n^6+120n^8+2n^{10})$
$\frac{3b_{3}-b_{1}}{b_{2}}=2n$
Then, a supposed recurrence : 
$\frac{1b_{2}-0b_{0}}{b_{1}}=n$
$\frac{2b_{4}-1b_{2}}{b_{3}}=n$
$\frac{3b_{6}-2b_{4}}{b_{5}}=n$
$\frac{4b_{8}-3b_{6}}{b_{7}}=n$
$\frac{5b_{10}-4b_{8}}{b_{9}}=n$
$...$
$\frac{k\,b_{2k}-(k-1)b_{2k-2}}{b_{2k-1}}=n$ , not proved.
A: Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series. 

We obtain
  \begin{align*}
\color{blue}{[x^j]\left(\frac{1+x}{1-x}\right)^n}
&=[x^j](1-x)^{-n}(1+x)^n\\
&=[x^j]\sum_{k=0}^\infty\binom{-n}{k}(-x)^k(1+x)^n\tag{1}\\
&=[x^j]\sum_{k=0}^\infty\binom{n+k-1}{k}x^k(1+x)^n\tag{2}\\
&=\sum_{k=0}^j\binom{n+k-1}{k}[x^{j-k}](1+x)^n\tag{3}\\
&\,\,\color{blue}{=\sum_{k=0}^j\binom{n+k-1}{k}\binom{n}{j-k}}\tag{4}
\end{align*}

Comment:


*

*In (1) we apply the binomial series expansion

*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$

*In (3) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the sum to $j$ since other values do not contribute.

*In (4) we select the coefficient of $[x^{j-k}](1+x)^n=[x^{j-k}]\sum_{t=0}^n\binom{n}{t}x^t$.
