Question

The term independent of x in the expansion of ${{\left( 1-x \right)}^{2}}.{{\left( x+\dfrac{1}{x} \right)}^{10}}$ is?

Answer 1

Use the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to expand the expression ${{\left( 1-x \right)}^{2}}$. Now, use the formula for the general term of the expression ${{\left( x+y \right)}^{n}}$ given as ${{T}_{r+1}}={}^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}$ and simplify the general term of the expression ${{\left( x+\dfrac{1}{x} \right)}^{10}}$. Select those values of r such that the product of the terms in the expansion of ${{\left( 1-x \right)}^{2}}$ with suitable terms of ${{\left( x+\dfrac{1}{x} \right)}^{10}}$ gives the terms independent of x. Take the sum of such coefficients and use the formula ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$ to get the answer.

Complete step by step answer:
Here we have been provided with the expression ${{\left( 1-x \right)}^{2}}.{{\left( x+\dfrac{1}{x} \right)}^{10}}$ and we are asked to find the term independent of x. Let us use some binomial of the expressions. Assuming the given expression as E we have,
$\Rightarrow E={{\left( 1-x \right)}^{2}}.{{\left( x+\dfrac{1}{x} \right)}^{10}}$
Using the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to expand the expression ${{\left( 1-x \right)}^{2}}$ we get,
$\Rightarrow E=\left( 1+{{x}^{2}}-2x \right).{{\left( x+\dfrac{1}{x} \right)}^{10}}$
Now, the general term of the expression ${{\left( x+y \right)}^{n}}$ is given as ${{T}_{r+1}}={}^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}$, so replacing y with $\dfrac{1}{x}$ and then using this formula for the general term of the expression ${{\left( x+\dfrac{1}{x} \right)}^{10}}$ we get,
$\begin{aligned} & \Rightarrow {{T}_{r+1}}={}^{10}{{C}_{r}}{{x}^{10-r}}{{\left( \dfrac{1}{x} \right)}^{r}} \\\ & \Rightarrow {{T}_{r+1}}={}^{10}{{C}_{r}}{{x}^{10-r}}{{x}^{-r}} \\\ & \Rightarrow {{T}_{r+1}}={}^{10}{{C}_{r}}{{x}^{10-2r}} \\\ \end{aligned}$
We need to take the terms independent of x, so we can think that if we multiply the 1 from the expression $\left( 1+{{x}^{2}}-2x \right)$ with the term that contains 0 as the exponent of x in the terms of ${{\left( x+\dfrac{1}{x} \right)}^{10}}$ we will get a term independent of x. So for the exponent of x to be zero in ${{\left( x+\dfrac{1}{x} \right)}^{10}}$ we must have the exponent of x in its general term equal to 0, so we get,
$\begin{aligned} & \Rightarrow 10-2r=0 \\\ & \Rightarrow r=5 \\\ \end{aligned}$
So, one of the terms is ${}^{10}{{C}_{5}}$.
Now, ${{x}^{2}}$ from $\left( 1+{{x}^{2}}-2x \right)$ must be multiplied with the term that contains -2 as the exponent of x in the terms of ${{\left( x+\dfrac{1}{x} \right)}^{10}}$, so we get,
$\begin{aligned} & \Rightarrow 10-2r=-2 \\\ & \Rightarrow r=6 \\\ \end{aligned}$
Therefore, one of the terms can also be ${}^{10}{{C}_{6}}$.
Finally, $-2x$ must be multiplied with the term containing -1 as the exponent of x in the terms of ${{\left( x+\dfrac{1}{x} \right)}^{10}}$, so we get,
$\begin{aligned} & \Rightarrow 10-2r=-1 \\\ & \Rightarrow r=\dfrac{11}{2} \\\ \end{aligned}$
But r cannot be fraction so the above value is neglected. Therefore, the required sum of the terms independent of x will be equal to ${}^{10}{{C}_{5}}+{}^{10}{{C}_{6}}$. Using the formula ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$ we get,
$\therefore {}^{10}{{C}_{5}}+{}^{10}{{C}_{6}}={}^{11}{{C}_{6}}$
Hence, the term independent of x in the expression ${{\left( 1-x \right)}^{2}}.{{\left( x+\dfrac{1}{x} \right)}^{10}}$ is ${}^{11}{{C}_{6}}$.

Note: Note that the fractional value of r is neglected because the combination formula is given as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ where r and r cannot be negative or fraction because the factorial of such numbers aren’t defined. Also, n must be greater than or equal to r. You can also write ${}^{11}{{C}_{6}}$ as ${}^{11}{{C}_{5}}$ because there is a property in combinations given as ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$.