Question

The term independent of $x$ in the expansion of ${\left[\frac{x+1}{x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+1}-\frac{x-1}{x-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right]}^{10}$, $x\ne 1$, is equal to

Answer 1

Step 1: Simplify the given expression:

Given expression is ${\left[\frac{x+1}{x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+1}-\frac{x-1}{x-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right]}^{10}$ and $x\ne 1$.

The above expression can be simplified as follows

$\begin{array}{cc}{\left[\frac{x+1}{x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+1}-\frac{x-1}{x-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right]}^{10}={\left[\frac{{\left(x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}^3+1^3}{x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+1}-\frac{{\left(x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)}^2-1^2}{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-1\right)}\right]}^{10}& \\ ={\left[\frac{\left(x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+1\right)\left(x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+1\right)}{\left(x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+1\right)}-\frac{\left(x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+1\right)\left(x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-1\right)}{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-1\right)}\right]}^{10}& \left[\because a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right),a^2-b^2=\left(a+b\right)\left(a-b\right)\right]\\ ={\left[x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+1-\left(\frac{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+1}{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)\right]}^{10}& \\ ={\left[x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+1-\frac{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{1}{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]}^{10}& \\ \therefore {\left[\frac{x+1}{x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+1}-\frac{x-1}{x-x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right]}^{10}={\left[x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-\frac{1}{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]}^{10}& \end{array}$

Step 2: Define the general term

From the binomial theorem, we know that,
${\left(a+b\right)}^n={}^{n}C_0{a}^n{b}^0+{}^{n}C_1{a}^{n-1}{b}^1+...+{}^{n}C_r{a}^{n-r}{b}^r+...+{}^{n}C_{n-1}{a}^1{b}^{n-1}+{}^{n}C_n{a}^0{b}^n$

The general term is,

$T_{r+1}={}^{n}C_r{a}^{n-r}{b}^r$
Here $n=10$, $a=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ and $b=\frac{1}{x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Thus,
$\begin{array}{rcl}{T}_{r+1}& =& {}^{10}C_r{\left(x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{10-r}{\left(\frac{1}{x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right)}^r\\ & =& {}^{10}C_r{\left(x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{10-r}{\left(x^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^r\\ & =& {}^{10}C_r{x}^{\frac{10-r}{3}-\frac{r}{2}}\end{array}$

Step 3: Set power of $x$ to $0$ in general term

For the term to be independent of $x$, the power of $x$ in the general term must equal $0$. Thus,
$\begin{array}{cc}& \frac{10-r}{3}-\frac{r}{2}=0\\ \Rightarrow & \frac{2\left(10-r\right)-3r}{6}=0\\ \Rightarrow & 20r-5r=0\\ \Rightarrow & r=4\end{array}$

Step 4: Calculate the coefficient for $r=4$

The coefficient when $r=4$ is, $T_{4+1}=T_5$ i.e., the fifth term, and its coefficient is,

${}^{10}C_4=\frac{10!}{\left(10-4\right)!\times 4!}=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}=210$

Therefore, the term that is independent of $x$ is $T_5$ and its coefficient is $210$.