Question

Coefficient of ${x^5}$ in the expansion of ${\left( {1 + {x^2}} \right)^5}{\left( {1 + x} \right)^4}$ is :
$
  {\text{A}}{\text{. 61}} \\
  {\text{B}}{\text{. 59}} \\
  {\text{C}}{\text{. 0}} \\
  {\text{D}}{\text{. 60}} \\
$

Answer 1

Hint: - To solve this question first we write expansion of both the terms then multiply as we can pick out the term required that means we will check where we are getting power of x is 5.

Complete step-by-step solution -
We should have knowledge of binomial expansion
${\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^n}{C_1}{a^{n - 1}}b{ + ^n}{C_2}{a^{n - 2}}{b^2} + \ldots \ldots { + ^n}{C_n}{a^0}{b^n}$
We have given,
${\left( {1 + {x^2}} \right)^5}{\left( {1 + x} \right)^4}$
$ \Rightarrow \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( {4{C_0} + 4{C_1}x + 4{C_2}{x^2} + 4{C_3}{x^3} + 4{C_4}{x^4}} \right)$
Now we have to use our brain to find the combination in which we will get power of x as 5.
So we have only two combinations where we are getting the required power of x.
$ \Rightarrow \left( {^5{C_1}{.^4}{C_3}{ + ^5}{C_2}{.^4}{C_1}} \right){x^5}$
So we have the required coefficient as
$\left( {^5{C_1}{.^4}{C_3}{ + ^5}{C_2}{.^4}{C_1}} \right)$
To solve this further we will use the formula $\left( {^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}} \right)$
On further calculation we get,
$ \Rightarrow 5.4 + \dfrac{{5 \times 4}}{2}.4 = 60$
Hence option D is the correct option.

Note:- Whenever we get this type of question the key concept of solving is first we have to remember ${\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^n}{C_1}{a^{n - 1}}b{ + ^n}{C_2}{a^{n - 2}}{b^2} + \ldots \ldots { + ^n}{C_n}{a^0}{b^n}$ to start the question then knowledge about $\left( {^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}} \right)$ this formula then we could be able to solve this type of question.