Question

find the coefficient of the term of ${{x}^{-5}}$in the binomial expansion of ${{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}$ , where $x\ne 0,1$.
(a) 1
(b) 4
(c) $-4$
(d) $-1$

Answer 1

In this question, we have to first find the binomial expansion of ${{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}$.
Using the formula of binomial expansion of elements say $a$ and $b$ raised to the power $n$ which is given by $\begin{aligned} & {{\left( a-b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}{{\left( -1 \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}{{\left( -1 \right)}^{r}}+... \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{\left( -1 \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}{{\left( -1 \right)}^{n}} \\\ \end{aligned}$
Where we have $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Also since the number of terms in the binomial expansion of ${{\left( a+b \right)}^{n}}$ is equal to $n+1$. Using this we will have that the number of terms in the binomial expansion of ${{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}$ is equals to 11. After finding the binomial expansion of ${{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}$we will have to determine the coefficient of the term of ${{x}^{-5}}$in the binomial expansion.

Complete step by step answer:
Let us first determine the binomial expansion of ${{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}$.
Since we know that the binomial expansion of ${{\left( a-b \right)}^{n}}$ raised to the power $n$ which is given by $\begin{aligned} & {{\left( a-b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}{{\left( -1 \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}{{\left( -1 \right)}^{r}}+... \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{\left( -1 \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}{{\left( -1 \right)}^{n}} \\\ \end{aligned}$Where we have $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
On comparing the expression ${{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}$ with ${{\left( a-b \right)}^{n}}$, we get that
$a=\dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}$, $b=\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}}$ and $n=10$.
We will now simplify the value of $a=\dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}$ using the identity that ${{x}^{3}}+{{y}^{3}}=\left( x+1 \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$.
Then we have

& a=\dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1} \\\ & =\dfrac{{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{3}}+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1} \\\ & =\dfrac{\left( {{x}^{\dfrac{1}{3}}}+1 \right)\left( {{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1 \right)}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1} \\\ & ={{x}^{\dfrac{1}{3}}}+1 \end{aligned}$$ That is we have $$a={{x}^{\dfrac{1}{3}}}+1$$. We will now simplify the value of $$b=\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}}$$ using the identity that $${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$$. Then we have $$\begin{aligned} & b=\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \\\ & =\dfrac{{{\left( {{x}^{2}} \right)}^{\dfrac{1}{2}}}-1}{x-{{x}^{\dfrac{1}{2}}}} \\\ & =\dfrac{\left( {{x}^{\dfrac{1}{2}}}+1 \right)\left( {{x}^{\dfrac{1}{2}}}-1 \right)}{{{x}^{\dfrac{1}{2}}}\left( {{x}^{\dfrac{1}{2}}}-1 \right)} \\\ & =\dfrac{{{x}^{\dfrac{1}{2}}}+1}{{{x}^{\dfrac{1}{2}}}} \\\ & =1+{{x}^{-\dfrac{1}{2}}} \end{aligned}$$ That is we have $$b=1+{{x}^{-\dfrac{1}{2}}}$$. Therefore the value of $$a-b$$ is given by $$\begin{aligned} & a-b={{x}^{\dfrac{1}{3}}}+1-\left( 1+{{x}^{-\dfrac{1}{2}}} \right) \\\ & ={{x}^{\dfrac{1}{3}}}+1-1-{{x}^{-\dfrac{1}{2}}} \\\ & ={{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \end{aligned}$$ Therefore we have simplified the binomial expansion of $${{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}$$ into $${{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}$$ We now have to expand the binomial expansion of $${{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}$$. Now using $$\begin{aligned} & {{\left( a-b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}{{\left( -1 \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}{{\left( -1 \right)}^{r}}+... \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{\left( -1 \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}{{\left( -1 \right)}^{n}} \\\ \end{aligned}$$Where we have $$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$. We will have $$\begin{aligned} & {{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}{{=}^{10}}{{C}_{0}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{10}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{10}}{{C}_{1}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{10-1}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{1}}{{\left( -1 \right)}^{1}}+...\,+ \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\,}^{10}}{{C}_{6}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{10-6}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{6}}{{\left( -1 \right)}^{6}}+...{{+}^{10}}{{C}_{9}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{1}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{9}}{{\left( -1 \right)}^{9}}{{+}^{10}}{{C}_{10}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{0}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{10}}{{\left( -1 \right)}^{10}} \\\ \end{aligned}$$ Now since we know that the number of terms in the binomial expansion of $${{\left( a-b \right)}^{n}}$$ is equals to $$n+1$$. Using this we will have that the number of terms in the binomial expansion of $${{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}$$ is equals to $$10+1=11$$ Also since by seeing the above binomial expansion, we have that the general term is given by $$^{10}{{C}_{r}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{r}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{10-r}}{{\left( -1 \right)}^{10-r}}{{=}^{10}}{{C}_{r}}{{\left( x \right)}^{\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)}}{{\left( -1 \right)}^{10-r}}..............(1)$$ Therefore in order to find coefficient of the term of $${{x}^{-5}}$$in the binomial expansion, we must have $$\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)=-5$$ Solving the above equation, we get $$\begin{aligned} & \dfrac{r}{3}+\dfrac{r}{2}-\dfrac{10}{2}=-5 \\\ & \Rightarrow \dfrac{r}{3}+\dfrac{r}{2}-5=-5 \\\ & \Rightarrow \dfrac{r}{3}+\dfrac{r}{2}=0 \\\ & \Rightarrow \dfrac{2r+3r}{6}=0 \\\ & \Rightarrow \dfrac{5r}{6}=0 \\\ & \Rightarrow r=0 \end{aligned}$$ Using substituting the value of $$r=0$$ in equation (1), we get $$^{10}{{C}_{r}}{{\left( x \right)}^{\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)}}{{\left( -1 \right)}^{10-r}}{{=}^{10}}{{C}_{0}}{{\left( x \right)}^{\dfrac{1}{3}\left( 0 \right)-\dfrac{1}{2}\left( 10-0 \right)}}{{\left( -1 \right)}^{10-0}}$$ Now since $$\begin{aligned} & ^{10}{{C}_{0}}=\dfrac{10!}{10!} \\\ & =1 \end{aligned}$$ Therefore we have $$\begin{aligned} & ^{10}{{C}_{r}}{{\left( x \right)}^{\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)}}{{\left( -1 \right)}^{10-r}}{{=}^{10}}{{C}_{0}}{{\left( x \right)}^{\dfrac{1}{3}\left( 0 \right)-\dfrac{1}{2}\left( 10-0 \right)}}{{\left( -1 \right)}^{10-0}} \\\ & =1\times {{x}^{-\dfrac{10}{2}}}\times 1 \\\ & ={{x}^{-5}} \end{aligned}$$ Therefore the coefficient of the term of $${{x}^{-5}}$$in the binomial expansion of $${{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}$$ is equals to 1. **So, the correct answer is “Option A”.** **Note:** In this problem, in order to determine the coefficient of the term of $${{x}^{-5}}$$in the binomial expansion of $${{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}$$ we have to find the binomial expansion of $${{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}$$ . Then carefully take the general term to find the coefficient.