Question

The number of terms independent of $x$ in the expansion of ${\left(x-\frac{1}{x}\right)}^4+{\left(x+\frac{1}{x}\right)}^3$ is

• A. $-3$
• B. $0$
• C. $3$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for correct option:
Given ${\left(x-\frac{1}{x}\right)}^4+{\left(x+\frac{1}{x}\right)}^3$

We know that ${\left(x+y\right)}^{\mathbf{n}}\mathbf{=}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{0}}\mathbf{·}{\mathbf{x}}^{\mathbf{n}}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{1}}\mathbf{·}{\mathbf{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{·}{\mathbf{y}}^{\mathbf{1}}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{2}}\mathbf{·}{\mathbf{x}}^{\mathbf{n}\mathbf{-}\mathbf{2}}\mathbf{·}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{·}{\mathbf{x}}^{\mathbf{1}}\mathbf{·}{\mathbf{y}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{n}}\mathbf{·}{\mathbf{y}}^{\mathbf{n}}$
Expansion of given expression
$$\begin{gathered} ={}^4{C}_0{\left(x\right)}^4.{\left(-\frac{1}{x}\right)}^0{+}^4{C}_1{\left(x\right)}^3.{\left(-\frac{1}{x}\right)}^1{+}^4{C}_2{\left(x\right)}^2.{\left(-\frac{1}{x}\right)}^2{+}^4{C}_3{\left(x\right)}^1.{\left(-\frac{1}{x}\right)}^3{+}^4{C}_4{\left(x\right)}^0.{\left(-\frac{1}{x}\right)}^4{+}^3{C}_0{\left(x\right)}^3.{\left(\frac{1}{x}\right)}^0{+}^3{C}_1{\left(x\right)}^2.{\left(\frac{1}{x}\right)}^1{+}^3{C}_2{\left(x\right)}^1.{\left(\frac{1}{x}\right)}^2{+}^3{C}_3{\left(x\right)}^0.{\left(\frac{1}{x}\right)}^3 \\ {=}^4{C}_0{\left(x\right)}^4{-}^4{C}_1{\left(x\right)}^2{+}^4{C}_2{-}^4{C}_3{\left(\frac{1}{x}\right)}^2{-}^4{C}_4{\left(\frac{1}{x}\right)}^4{+}^3{C}_0{\left(x\right)}^3{+}^3{C}_1\left(x\right){+}^3{C}_2\left(\frac{1}{x}\right){+}^3{C}_3{\left(\frac{1}{x}\right)}^3 \end{gathered}$$Therefore only one term is independent of $x$ i.e. ${}^{4}C_2$

Hence option(D) is correct.