Question

The sum of the coefficients of all odd degree terms in the expansion of ${\left(x+\sqrt{\left(x^3-1\right)}\right)}^5+{\left(x-\sqrt{\left(x^3-1\right)}\right)}^5$ is

• A. $1$
• B. $2$
• C. $-1$
• D. $0$

Answer 1

The correct option is B

$2$

Explanation for the correct answer:

By using the Binomial theorem we know that,

${\left(a+b\right)}^n={}^n{C}_0{a}^n+{}^n{C}_1{a}^{n-1}b+{}^n{C}_2{a}^{n-2}{b}^2+......+{}^n{C}_n{b}^n$ $...\left(i\right)$

${\left(a-b\right)}^n={}^n{C}_0{a}^n-{}^n{C}_1{a}^{n-1}b+{}^n{C}_2{a}^{n-2}{b}^2-......+{\left(-1\right)}^n{}^n{C}_n{b}^n$ $...\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$ we get,

${\left(a+b\right)}^n+{\left(a-b\right)}^n=2\left({}^n{C}_0{a}^n+{}^n{C}_2{a}^{n-2}{b}^2+{}^n{C}_4{a}^{n-4}{b}^4+.....\right)$ $...\left(iii\right)$

Now in equation $\left(iii\right)$ if we substitute $a=x,b=\sqrt{x^3-1},n=5$, we get

${\left(x+\sqrt{\left(x^3-1\right)}\right)}^5+{\left(x-\sqrt{\left(x^3-1\right)}\right)}^5=2\left({}^{5}C_0{x}^5+{}^{5}C_2{x}^3\left(x^3-1\right)+{}^{5}C_4{x}^1{\left(x^3-1\right)}^2\right)$

$=2\left(\frac{5!}{0!5!}{x}^5+\frac{5!}{2!3!}\left(x^6-x^3\right)+\frac{5!}{4!1!}\left(x^7-2x^4+x\right)\right)$

$=2\left(x^5+10\left(x^6-x^3\right)+5\left(x^7-2x^4+x\right)\right)$

$\Rightarrow$ ${\left(x+\sqrt{\left(x^3-1\right)}\right)}^5+{\left(x-\sqrt{\left(x^3-1\right)}\right)}^5$$=2\left(5x-10x^3-10x^4+x^5+10x^6+5x^7\right)$

Hence the sum of odd degree coefficients $=2\left(5-10+1+5\right)$

$$\begin{gathered} =2\left(1\right) \\ =2 \end{gathered}$$

Hence, the sum of the odd degree coefficients of ${\left(x+\sqrt{\left(x^3-1\right)}\right)}^5+{\left(x-\sqrt{\left(x^3-1\right)}\right)}^5$ is $2$.

Hence, option B is the correct answer.