Question

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

Answer 1

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.

Complete step-by-step answer:
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.