<s>[INST] 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.
 [/INST] 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 $[x^{j-k}](1+x)^n=[x^{j-k}]\sum_{t=0}^n\binom{n}{t}x^t$.
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