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Question: The sum of coefficients of all odd degree terms in the expansion of \({\left( {x + \sqrt {{x^3} - 1}...

The sum of coefficients of all odd degree terms in the expansion of (x+x3−1)5+(x−x3−1)5, (x>1){\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5},{\text{ }}\left( {x > 1} \right) is
A) 11
B) 22
C) −1 - 1
D) 00

Explanation

Solution

We know how to expand a binomial expression like
(a+b)n=nC0an+nC1an−1b+nC2an−2b2+ - - - - - - - - - - - - - - - - - - nCna0bn{\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
(x+x3−1)5+(x−x3−1)5{\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}
So, let us use binomial expansion formula
(a+b)n=nC0an+nC1an−1b+nC2an−2b2+ - - - - - - - - - - - - - - - - - - nCna0bn{\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 (x+x3−1)5{\left( {x + \sqrt {{x^3} - 1} } \right)^5}
So, here a=x and b=x3−1a = x{\text{ and }}b = \sqrt {{x^3} - 1} .
So, (x+x3−1)5=x5+5C1x4x3−1+5C2x3(x3−1)+5C3x2(x3−1)x3−1+5C4x(x3−1)2+5C5(x3−1)2x3−1{\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,
(x+x3−1)5=x5+5!4!1!x4x3−1+5!2!3!x3(x3−1)+5!2!3!x2(x3−1)x3−1+5!4!1!x(x3−1)2+5!5!0!(x3−1)2x3−1 =x5+5x4x3−1+10x6−10x3+10x5x3−1−10x2x3−1+5x(x6+1−2x3)+(x6+1−2x3)x3−1 - - - - - - - - (1)  {\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 (x−x3−1)5{\left( {x - \sqrt {{x^3} - 1} } \right)^5},
So, (a−b)n{\left( {a - b} \right)^n}. So, it has two cases.
If n=evenn = {\text{even}}, then
an−nC1an−1b+nC2an−2b2+ - - - - - - - - - - - - - - - - - - + bn{a^n} - {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + {\text{ - - - - - - - - - - - - - - - - - - + }}{b^n}
If n=oddn = {\text{odd}}, then
an−nC1an−1b+nC2an−2b2+ - - - - - - - - - - - - - - - - - - - bn{a^n} - {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + {\text{ - - - - - - - - - - - - - - - - - - - }}{b^n}
Here, n=5n = 5, that is odd, so,
(x−x3−1)5=x5−5C1x4x3−1+5C2x3(x3−1)−5C3x2(x3−1)x3−1+5C4x(x3−1)2−5C5(x3−1)2x3−1{\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,
(x−x3−1)5=x5−5!4!1!x4x3−1+5!2!3!x3(x3−1)−5!2!3!x2(x3−1)x3−1+5!4!1!x(x3−1)2−5!5!0!(x3−1)2x3−1 =x5−5x4x3−1+10x6−10x3−10x5x3−1+10x2x3−1+5x(x6+1−2x3)−(x6+1−2x3)x3−1 - - - - - - - - (2)  {\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 (x+x3−1)5+(x−x3−1)5=S{\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5} = S
We will get
⇒S=(2x5+20x6−20x3+10x(x6+1−2x3)) ⇒S=(2x5+20x6−20x3+10x7+10x−20x4)  \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−20x3−20x4+2x5+20x6+10x7S = 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 = 10 - 20 + 2 + 10 = 2
So, our answer is 22.

So, option B is the correct answer.

Note: We can do it by alternative method, that is, if (A+B)n+(A−B)n{\left( {A + B} \right)^n} + {\left( {A - B} \right)^n} is given, then its expansion will be
2[An+nC2An−2B2+nC4An−4B4 - - - - - - - - ]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.