Question

The sum of coefficients of all odd degree terms in the expansion of ${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5},{\text{ }}\left( {x > 1} \right)$ is
A) $1$
B) $2$
C) $ - 1$
D) $0$

Answer 1

Hint: We know how to expand a binomial expression like
${\left( {a + b} \right)^n} = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + {\text{ - - - - - - - - - - - - - - - - - - }}{}^n{C_n}{a^0}{b^n}$
Similarly, expand both the expression and add them , then we will find an expression from which you need to add the coefficient of all odd degree terms.

Complete step-by-step answer:
Here, according to question, we are given an expression
${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}$
So, let us use binomial expansion formula
${\left( {a + b} \right)^n} = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + {\text{ - - - - - - - - - - - - - - - - - - }}{}^n{C_n}{a^0}{b^n}$
So, let us first expand ${\left( {x + \sqrt {{x^3} - 1} } \right)^5}$
So, here $a = x{\text{ and }}b = \sqrt {{x^3} - 1} $.
So, ${\left( {x + \sqrt {{x^3} - 1} } \right)^5} = {x^5} + {}^5{C_1}{x^4}\sqrt {{x^3} - 1} + {}^5{C_2}{x^3}\left( {{x^3} - 1} \right) + {}^5{C_3}{x^2}\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} + {}^5{C_4}x{\left( {{x^3} - 1} \right)^2} + {}^5{C_5}{\left( {{x^3} - 1} \right)^2}\sqrt {{x^3} - 1} $
On further expansion we get,
$
  {\left( {x + \sqrt {{x^3} - 1} } \right)^5} = {x^5} + \dfrac{{5!}}{{4!1!}}{x^4}\sqrt {{x^3} - 1} + \dfrac{{5!}}{{2!3!}}{x^3}\left( {{x^3} - 1} \right) + \dfrac{{5!}}{{2!3!}}{x^2}\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} + \dfrac{{5!}}{{4!1!}}x{\left( {{x^3} - 1} \right)^2} + \dfrac{{5!}}{{5!0!}}{\left( {{x^3} - 1} \right)^2}\sqrt {{x^3} - 1} \\
   = {x^5} + 5{x^4}\sqrt {{x^3} - 1} + 10{x^6} - 10{x^3} + 10{x^5}\sqrt {{x^3} - 1} - 10{x^2}\sqrt {{x^3} - 1} + 5x\left( {{x^6} + 1 - 2{x^3}} \right) + \left( {{x^6} + 1 - 2{x^3}} \right)\sqrt {{x^3} - 1} {\text{ - - - - - - - - (1)}} \\
 $
Now upon expanding ${\left( {x - \sqrt {{x^3} - 1} } \right)^5}$,
So, ${\left( {a - b} \right)^n}$. So, it has two cases.
If $n = {\text{even}}$, then
${a^n} - {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + {\text{ - - - - - - - - - - - - - - - - - - + }}{b^n}$
If $n = {\text{odd}}$, then
${a^n} - {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + {\text{ - - - - - - - - - - - - - - - - - - - }}{b^n}$
Here, $n = 5$, that is odd, so,
${\left( {x - \sqrt {{x^3} - 1} } \right)^5} = {x^5} - {}^5{C_1}{x^4}\sqrt {{x^3} - 1} + {}^5{C_2}{x^3}\left( {{x^3} - 1} \right) - {}^5{C_3}{x^2}\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} + {}^5{C_4}x{\left( {{x^3} - 1} \right)^2} - {}^5{C_5}{\left( {{x^3} - 1} \right)^2}\sqrt {{x^3} - 1} $
On further expanding,
$
  {\left( {x - \sqrt {{x^3} - 1} } \right)^5} = {x^5} - \dfrac{{5!}}{{4!1!}}{x^4}\sqrt {{x^3} - 1} + \dfrac{{5!}}{{2!3!}}{x^3}\left( {{x^3} - 1} \right) - \dfrac{{5!}}{{2!3!}}{x^2}\left( {{x^3} - 1} \right)\sqrt {{x^3} - 1} + \dfrac{{5!}}{{4!1!}}x{\left( {{x^3} - 1} \right)^2} - \dfrac{{5!}}{{5!0!}}{\left( {{x^3} - 1} \right)^2}\sqrt {{x^3} - 1} \\
   = {x^5} - 5{x^4}\sqrt {{x^3} - 1} + 10{x^6} - 10{x^3} - 10{x^5}\sqrt {{x^3} - 1} + 10{x^2}\sqrt {{x^3} - 1} + 5x\left( {{x^6} + 1 - 2{x^3}} \right) - \left( {{x^6} + 1 - 2{x^3}} \right)\sqrt {{x^3} - 1} {\text{ - - - - - - - - (2)}} \\
 $
So, upon adding equation (1) and (2), we get
Let ${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5} = S$
We will get
$
\Rightarrow S = \left( {2{x^5} + 20{x^6} - 20{x^3} + 10x\left( {{x^6} + 1 - 2{x^3}} \right)} \right) \\
\Rightarrow S = \left( {2{x^5} + 20{x^6} - 20{x^3} + 10{x^7} + 10x - 20{x^4}} \right) \\
 $
Now, on rearranging,
$\Rightarrow$$S = 10x - 20{x^3} - 20{x^4} + 2{x^5} + 20{x^6} + 10{x^7}$
Now, sum of coefficient of odd power that means$ = 10 - 20 + 2 + 10 = 2$
So, our answer is $2$.

So, option B is the correct answer.

Note: We can do it by alternative method, that is, if ${\left( {A + B} \right)^n} + {\left( {A - B} \right)^n}$ is given, then its expansion will be
$2\left[ {{A^n} + {}^n{C_2}{A^{n - 2}}{B^2} + {}^n{C_4}{A^{n - 4}}{B^4}{\text{ - - - - - - - - }}} \right]$
We can directly use this formula and calculate the results.