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

Hint: 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 \[{}^n{C_k}\].