Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The sum of coefficients of all even degree terms in $x$ in the expansion of ${\left( {x + \sqrt {{x^3} - 1} } \right)^6} + {\left( {x - \sqrt {{x^3} - 1} } \right)^6},\left( {x > 1} \right)$ isA.32B.26C.29D.24

Last updated date: 08th Sep 2024
Total views: 427.8k
Views today: 9.27k
Verified
427.8k+ views
Hint: Here, we will use the formula,${\left( {a + b} \right)^n} + {\left( {a - b} \right)^n} = 2\left( {{}^n{C_0}{a^n} + {}^n{C_2}{a^{n - 2}} + {}^n{C_4}{a^{n - 4}}{b^4} + ...} \right)$ in the given expression and then we will use formula to calculate combinations is ${}^n{C_r} = \dfrac{{\left. {\underline {\, n \,}}\! \right| }}{{\left. {\underline {\, r \,}}\! \right| \cdot \left. {\underline {\, {n - r} \,}}\! \right| }}$, where $n$ is the number of items, and $r$ represents the number of items being chosen, in the equation. Then we will simplify to find the required value.

We are given that the expression is ${\left( {x + \sqrt {{x^3} - 1} } \right)^6} + {\left( {x - \sqrt {{x^3} - 1} } \right)^6},\left( {x > 1} \right)$.
We will use the formula,${\left( {a + b} \right)^n} + {\left( {a - b} \right)^n} = 2\left( {{}^n{C_0}{a^n} + {}^n{C_2}{a^{n - 2}} + {}^n{C_4}{a^{n - 4}}{b^4} + ...} \right)$ in the given expression.
Finding the value of $a$, $b$ and $n$ from the given expression, we get
$a = x$
$b = \sqrt {{x^3} - 1}$
$n = 6$
Using the above values in the above formula, we get
$\Rightarrow 2\left( {{}^6{C_0}{x^6} + {}^6{C_2}{x^{6 - 2}}\left( {{x^3} - 1} \right) + {}^6{C_4}{x^{6 - 4}}{{\left( {{x^3} - 1} \right)}^2} + {}^6{C_6}{{\left( {{x^3} - 1} \right)}^3}} \right) \\ \Rightarrow 2\left( {{}^6{C_0}{x^6} + {}^6{C_2}{x^4}\left( {{x^3} - 1} \right) + {}^6{C_4}{x^2}{{\left( {{x^3} - 1} \right)}^2} + {}^6{C_6}{{\left( {{x^3} - 1} \right)}^3}} \right) \\ \Rightarrow 2\left( {{}^6{C_0}{x^6} + {}^6{C_2}{x^7} - {}^6{C_2}{x^4} + {}^6{C_4}{x^2}\left( {{x^6} + 1 - 2{x^3}} \right) + {}^6{C_6}\left( {{x^9} - 1 - 3{x^6} + 3{x^3}} \right)} \right) \\ \Rightarrow 2\left( {{}^6{C_0}{x^6} + {}^6{C_2}{x^7} - {}^6{C_2}{x^4} + {}^6{C_4}{x^8} + {}^6{C_4}{x^2} - {}^6{C_4}2{x^5} + {}^6{C_6}{x^9} - {}^6{C_6} - {}^6{C_6}3{x^6} + {}^6{C_6}3{x^3}} \right) \\$

Using formula to calculate combinations is ${}^n{C_r} = \dfrac{{\left. {\underline {\, n \,}}\! \right| }}{{\left. {\underline {\, r \,}}\! \right| \cdot \left. {\underline {\, {n - r} \,}}\! \right| }}$, where $n$ is the number of items, and $r$ represents the number of items being chosen, in the above equation, we get
$\Rightarrow 2\left( {\dfrac{{6!}}{{6!}}{x^6} + \dfrac{{6!}}{{2!4!}}{x^7} - \dfrac{{6!}}{{2!4!}}{x^4} + \dfrac{{6!}}{{4!2!}}{x^8} + \dfrac{{6!}}{{4!2!}}{x^2} - \dfrac{{6!}}{{4!2!}}2{x^5} + {x^9} - 1 - 3{x^6} + 3{x^3}} \right) \\ \Rightarrow 2\left( {{x^6} + 6{x^7} - 15{x^4} + 15{x^8} + 15{x^2} - 30{x^5} + {x^9} - 1 - 3{x^6} + 3{x^3}} \right) \\$
Finding the sum of coefficient of even powers from the above expression, we get
$\Rightarrow 2\left[ {1 - 15 + 15 + 15 - 1 - 3} \right] \\ \Rightarrow 2\left[ {12} \right] \\ \Rightarrow 24 \\$
Therefore, sum of coefficients of all even degree terms in $x$ in the expansion of ${\left( {x + \sqrt {{x^3} - 1} } \right)^6} + {\left( {x - \sqrt {{x^3} - 1} } \right)^6},\left( {x > 1} \right)$ is 24.
Hence, option D is correct.

Note: In solving these types of questions, you should be familiar with the formula of the area of the rectangle. Then use the given conditions and values given in the question, and substitute in the formula, to find the required values. Also, we are supposed to write the values properly to avoid any miscalculation.