Question

The sum of the co-efficients of all odd degree terms in the expansion of ${(x+\sqrt{x^3-1})}^5+(x-\sqrt{x^3-1}{)}^5,(x>1)$ is

• A. $1$
• B. $2$
• C. $-1$
• D. $0$

Answer 1

Answer: (B)

$2$

$(x+\sqrt{x^3-1}{)}^5+(x-\sqrt{x^3-1}{)}^5$

$(a+b{)}^5=a^5+5a^{4}b+10a^3{b}^2+10a^2{b}^3+5a{b}^4+b^5-------(i)$

$(a-b{)}^5=a^5-5a^{4}b+10a^3{b}^2-10a^2{b}^3+5a{b}^4-b^5-------(ii)$

$(a+b{)}^5+(a-b{)}^5=2[a^5+10a^3{b}^2+5a{b}^4]$

$=2[x^5+10x^3(x^3-1)+5x(x^3-1{)}^4]$

$=2[x^5+10x^6-10x^3+5x(x^6-2x^3+1\left)\right]$

$=2x^5+20x^6-20x^3+10x^7-20x^4+10x$

Here all the co-efficient of the above equation are odd terms coefficients of even terms are cancelled out.

So, sum of coefficients $\Rightarrow 2+20-20+10-20+10=2$