Question

The term independent of x in the binomial expansion of
$\left( {1 - \dfrac{1}{x} + 3{x^5}} \right){\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ is
$\left( A \right)$ - 496
$\left( B \right)$ - 400
$\left( C \right)$ 496
$\left( D \right)$ 400

Answer 1

Hint – In this particular type of question use the concept that according to Binomial expansion, the expansion of, ${\left( {b - a} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{b^{n - r}}{{\left( { - a} \right)}^r}}$and later on use the concept that the independent term of x in the Binomial expansion of the given equation is the sum of product of individual terms which cancel out all the variable so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation:
$\left( {1 - \dfrac{1}{x} + 3{x^5}} \right){\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$
Now when we multiply ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ to $\left( {1 - \dfrac{1}{x} + 3{x^5}} \right)$ there are three terms which is independent of x, which is given as, constant term (i.e. independent of x), coefficient of x and coefficient of $\dfrac{1}{{{x^5}}} = {x^{ - 5}}$ in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ using Binomial theorem.
So when we multiply these three terms with the individual terms of $\left( {1 - \dfrac{1}{x} + 3{x^5}} \right)$, then we get the required term independent of x in the binomial expansion.
So, first out these three terms in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$.
As we know according to Binomial expansion, the expansion of
${\left( {b - a} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{b^{n - r}}{{\left( { - a} \right)}^r}}$
So, on comparing $b = 2{x^2},{\text{ }}a = \dfrac{1}{x},{\text{ }}n = 8$
$\Rightarrow {\left( {2{x^2} - \dfrac{1}{x}} \right)^8} = \sum\limits_{r = 0}^8 {{}^8{C_r}{{\left( {2{x^2}} \right)}^{8 - r}}{{\left( { - \dfrac{1}{x}} \right)}^r}}$
$= \sum\limits_{r = 0}^8 {{}^8{C_r}{{\left( x \right)}^{16 - 2r}}{{\left( 2 \right)}^{8 - r}}{{\left( { - 1} \right)}^r}{{\left( x \right)}^{ - r}}} = \sum\limits_{r = 0}^8 {{}^8{C_r}{{\left( x \right)}^{16 - 3r}}{{\left( { - 1} \right)}^r}{{\left( 2 \right)}^{8 - r}}}$
Now first find out the constant term.
So, put the power of $x$in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ equal to zero.
$\Rightarrow 16 - 3r = 0 \\\ \Rightarrow 3r = 16 \\\ \Rightarrow r = \dfrac{{16}}{3} \\\$
So, r = (16/3) is not possible, so there is no term independent of x in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$
Now find out the coefficient of x.
So, put the power of $x$in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ equal to one.
$\Rightarrow 16 - 3r = 1 \\\ \Rightarrow 3r = 15 \\\ \Rightarrow r = \dfrac{{15}}{3} = 5 \\\$
So the term of x in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ = ${}^8{C_5}{\left( x \right)^{16 - 3\left( 5 \right)}}{\left( { - 1} \right)^5}{\left( 2 \right)^{8 - 5}}$
So the coefficient of x in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ = ${}^8{C_5}{\left( { - 1} \right)^5}{\left( 2 \right)^{8 - 5}}$
Now find out the term of $\dfrac{1}{{{x^5}}} = {x^{ - 5}}$ in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$.
So, put the power of $x$in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ equal to (-5).
$\Rightarrow 16 - 3r = - 5 \\\ \Rightarrow 3r = 21 \\\ \Rightarrow r = \dfrac{{21}}{3} = 7 \\\$
So the term of ($\dfrac{1}{{{x^5}}} = {x^{ - 5}}$) in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ = ${}^8{C_7}{\left( x \right)^{16 - 3\left( 7 \right)}}{\left( { - 1} \right)^7}{\left( 2 \right)^{8 - 7}}$
So the coefficient of ($\dfrac{1}{{{x^5}}} = {x^{ - 5}}$) in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ = ${}^8{C_7}{\left( { - 1} \right)^7}{\left( 2 \right)^{8 - 7}}$
So, the term independent of x in the binomial expansion is given as,
$\Rightarrow \left( {1 - \dfrac{1}{x} + 3{x^5}} \right){\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ = (-1) times the coefficient of x in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ + (3) times the coefficient of ($\dfrac{1}{{{x^5}}} = {x^{ - 5}}$) in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$.
$\Rightarrow \left( {1 - \dfrac{1}{x} + 3{x^5}} \right){\left( {2{x^2} - \dfrac{1}{x}} \right)^8} = \left( { - 1} \right)\left[ {{}^8{C_5}{{\left( { - 1} \right)}^5}{{\left( 2 \right)}^{8 - 5}}} \right] + 3\left[ {{}^8{C_7}{{\left( { - 1} \right)}^7}{{\left( 2 \right)}^{8 - 7}}} \right]$
Now simplify this we have,
$\Rightarrow \left( {1 - \dfrac{1}{x} + 3{x^5}} \right){\left( {2{x^2} - \dfrac{1}{x}} \right)^8} = \left[ {{}^8{C_5}{{\left( 2 \right)}^3}} \right] - 3\left[ {{}^8{C_7}\left( 2 \right)} \right]$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so we have,
$\Rightarrow \left( {1 - \dfrac{1}{x} + 3{x^5}} \right){\left( {2{x^2} - \dfrac{1}{x}} \right)^8} = \left[ {\dfrac{{8!}}{{5!\left( {8 - 5} \right)!}}8} \right] - 6\left[ {\dfrac{{8!}}{{7!\left( {8 - 7} \right)!}}} \right]$
$\Rightarrow \left( {1 - \dfrac{1}{x} + 3{x^5}} \right){\left( {2{x^2} - \dfrac{1}{x}} \right)^8} = \left[ {\dfrac{{8.7.6.5!}}{{5!.3.2.1}}8} \right] - 6\left[ {\dfrac{{8.7!}}{{7!}}} \right]$
$\Rightarrow \left( {1 - \dfrac{1}{x} + 3{x^5}} \right){\left( {2{x^2} - \dfrac{1}{x}} \right)^8} = 8.7.8 - 6.8 = 448 - 48 = 400$
So this is the required answer.
Hence option (D) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the combination (i.e.${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$), then first find out all the terms in the expansion of ${\left( {2{x^2} - \dfrac{1}{x}} \right)^8}$ such that when these terms is multiplied by the terms of $\left( {1 - \dfrac{1}{x} + 3{x^5}} \right)$ we get the terms which is independent of x, as above then simplify as above we get the required answer.