Question

The coefficient of $x^5$ in the expansion of ${(1+x^2)}^5{(1+x)}^4$ is:

• A. $30$
• B. $60$
• C. $40$
• D. None of the these

Answer 1

The correct option is B

$60$

Explanation for the correct option:

Finding the coefficient of $x^5$:

Given expression is ${(1+x^2)}^5{(1+x)}^4$

We know that binomial expansion of ${\left(a+b\right)}^{\mathbf{n}}\mathbf{=}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{0}}{\mathbf{a}}^{\mathbf{n}}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{1}}{\mathbf{a}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{b}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{2}}{\mathbf{a}}^{\mathbf{n}\mathbf{-}\mathbf{2}}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}\mathbf{\cdots }\mathbf{\cdots }\mathbf{\cdots }\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{n}}{\mathbf{a}}^{\mathbf{1}}{\mathbf{b}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{+}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{n}}{\mathbf{b}}^{\mathbf{n}}$

Simplifying the expression using binomial expansion

${(1+x^2)}^5{(1+x)}^4=\left({}^{5}C_0+{}^{5}C_1{x}^2+{}^{5}C_2{x}^4+{}^{5}C_3{x}^6+{}^{5}C_4{x}^8+{}^{5}C_5{x}^{10}\right)\left(1+{}^{4}C_{1}x+{}^{4}C_2{x}^2+{}^{4}C_3{x}^3+{}^{4}C_4{x}^4\right)$

Here, $x^5$ is obtained by multiplication of $x^2$ and $x^3$ and also by multiplication of $x^4$ and $x^1$.

Therefore, the coefficient of $x^5$ is:

$$\begin{gathered} ={}^{5}C_1{}^{4}C_3+{}^{5}C_2{}^{4}C_1 \\ =\frac{5!}{4!}.\frac{4!}{3!}+\frac{5!}{2!.3!}.\frac{4!}{3!} \\ =\frac{5\times 4!}{4!}.\frac{4\times 3!}{3!}+\frac{5\times 4\times 3\times 2\times 1}{1\times 2\times 1\times 2\times 3}.\frac{4\times 3!}{3!} \\ =\left(5\times 4\right)+\left(10\times 4\right) \\ =20+40 \\ =60 \end{gathered}$$

Hence, option (B) is the correct answer.