Question

Find the correct option of the coefficient of ${x^2}$ in the expansion of the product $\left( {2 - {x^2}} \right) \cdot \left( {{{\left( {1 + 2x + 3{x^2}} \right)}^6} + {{\left( {1 - 4{x^2}} \right)}^6}} \right)$.
A. 106
B. 107
C. 155
D. 108

Answer 1

First we will multiply the expression $\left( {{{\left( {1 + 2x + 3{x^2}} \right)}^6} + {{\left( {1 - 4{x^2}} \right)}^6}} \right)$ with outer term $\left( {2 - {x^2}} \right)$ and using binomial expansion expand the terms and then separate the terms which are having the coefficient of ${x^2}$. Use the formal expression of the binomial theorem, ${\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {{}^n{C_k}{a^{n - k}}{b^k}}$ to expand the value of the given expression.

Complete step by step solution:
First consider the formula which we will used to simplify,

$${\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {{}^n{C_k}{a^{n - k}}{b^k}} \\\ = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + .... \\\$$

First, we will expand the term ${\left( {1 + 2x + 3{x^2}} \right)^6}$ using the binomial formula written above.
We get,

$${\left( {1 + \left( {2x + 3{x^2}} \right)} \right)^6} = {}^6{C_0}{\left( 1 \right)^6} + {}^6{C_1}{\left( 1 \right)^{6 - 1}}{\left( {2x + 3{x^2}} \right)^1} + {}^6{C_2}{\left( 1 \right)^{6 - 2}}{\left( {2x + 3{x^2}} \right)^2} + ... \\\ = \left[ {1\left( 1 \right) + 1\left( {2x + 3{x^2}} \right)6 + 15\left( {4{x^2} + 9{x^4} + 12{x^3}} \right)} \right] + .... \\\$$

Now, multiply the outer term $\left( {2 - {x^2}} \right)$ with the values obtained above.
Thus, we get,

$$\left( {2 - {x^2}} \right){\left( {\left( {1 + \left( {2x + 3{x^2}} \right)} \right)} \right)^6} = \left( {2 - {x^2}} \right)\left[ {1\left( 1 \right) + 1\left( {2x + 3{x^2}} \right)6 + 15\left( {4{x^2} + 9{x^4} + 12{x^3}} \right)} \right] + .... \\\ = \left( {2 - {x^2}} \right) + \left( {2 - {x^2}} \right)\left( {2x + 3{x^2}} \right)6 + 15\left( {2 - {x^2}} \right)\left( {4{x^2} + 9{x^4} + 12{x^3}} \right) + ... \\\ = 2 - {x^2} + \left( {4x + 6{x^2} - 2{x^3} - 3{x^4}} \right)6 + 15\left( {8{x^2} + 18{x^4} + 24{x^3} - 4{x^4} - 9{x^6} - 12{x^5}} \right) + … \\\$$

Other higher terms in the expansion will contain terms with coefficient higher than ${x^2}$. So we’ll neglect those terms.
Now, we will consider the terms which are having ${x^2}$. Thus we get,
$\Rightarrow - {x^2} + 36{x^2} + 120{x^2} = 155{x^2}$ -----(i) expression
Next, we will expand the term ${\left( {1 - 4{x^2}} \right)^6}$ using the binomial formula
We get,

$${\left( {1 - 4{x^2}} \right)^6} = {}^6{C_0}{\left( 1 \right)^6} + {}^6{C_1}{\left( 1 \right)^{6 - 1}}{\left( { - 4{x^2}} \right)^1} + {}^6{C_2}{\left( 1 \right)^{6 - 2}}{\left( { - 4{x^2}} \right)^2} + …\\\ = \left[ {1 + 1\left( 6 \right)\left( { - 4{x^2}} \right) + 15\left( { - 4{x^4}} \right)} \right] + …\\\$$

Neglecting other higher terms.
Now, multiply the outer term $\left( {2 - {x^2}} \right)$ with the values obtained above.
Thus, we get,

$$\left( {2 - {x^2}} \right){\left( {1 - 4{x^2}} \right)^6} = \left( {2 - {x^2}} \right)\left[ {1 - 24{x^2}} \right] \\\ = 2 - {x^2} - 48{x^2} + 24{x^4} \\\ = 2 - 49{x^2} + 24{x^4} \\\$$

Now, we will consider the terms which are having ${x^2}$. Thus we get,
$\Rightarrow - 49{x^2}$ -------(ii) expression
Now, from the first expression we can see that the coefficient of ${x^2}$ is 155 and from the second expression we have coefficient of ${x^2}$ as $- 49$
Thus, now, we will calculate the total of both the coefficients by adding their coefficients obtained above.
Hence, we get,
$155 - 49 = 106$
Thus, the coefficient of ${x^2}$ is 106.
Thus, option A is correct.
Note: We have separated the terms in the beginning only by multiplying and then applying the binomial formula. Ignore those terms whose coefficient is not ${x^2}$ and choose only those terms who is having the term ${x^2}$. We have expanded the value of ${}^n{C_r}$ as this ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ to evaluate the values of ${}^n{C_k}$.