Question

The term independent of $x$ in the expansion of ${(1+2x+\frac{2}{x})}^3$ is.........

• A. $25$
• B. $8$
• C. $14$
• D. $Noneofthese$

Answer 1

Answer: (A)

$25$

$\begin{array}{c}{\left(1+2x+\frac{2}{x}\right)}^3\\ \Rightarrow {\left\{1+\left(2x+\frac{2}{x}\right)\right\}}^3\end{array}$

By using binomial expansion

$\begin{array}{c}{\Rightarrow }^3{c}_0{\left(2x+\frac{2}{x}\right)}^3{+}^3{c}_1{\left(2x+\frac{2}{x}\right)}^2{+}^3{c}_2\left(2x+\frac{2}{x}\right){+}^3{c}_3+1\\ \Rightarrow {\left(2x+\frac{2}{x}\right)}^3+3{\left(2x+\frac{2}{x}\right)}^2+3\left(2x+\frac{2}{x}\right)+1\end{array}$

Again by using binomial expansion

$\begin{array}{c}\Rightarrow 2^3{\left(x+\frac{1}{x}\right)}^3+12{\left(x+\frac{1}{x}\right)}^2+6\left(x+\frac{1}{x}\right)+1\\ \Rightarrow 8\left[{}^3{c}_0.x^{\circ }-\frac{1}{x^3}{+}^3{c}_1,x.\frac{1}{x}{+}^3{c}_2{x}^2\frac{1}{x}\right]+12\left[{}^2{c}_0{x}^{\circ }.\frac{1}{x^2}{+}^2{c}_{1}x.\frac{1}{x}\right]+6x+\frac{6}{x}+1\end{array}$

Term independent of $x$ is $8{\times }^3{c}_2+12{\times }^2{c}_1+1=\left(+24+1\right)=25.$

Hence, this is the answer.