Question:
The sum of the co-efficients of all even degree terms in $x$ in the expansion of $\left(x+\sqrt{x^{3}-1}\right)^{6}+\left(x-\sqrt{x^{3}-1}\right)^{6},(x>1)$ is equal to :
Correct Option: , 4
Solution:
$\left(x+\sqrt{x^{3}-1}\right)^{6}+\left(x-\sqrt{x^{3}-1}\right)^{6}$
$=2\left[{ }^{6} C_{0} x^{6}+{ }^{6} C_{2} x^{4}\left(x^{3}-1\right)+{ }^{6} C_{4} x^{2}\left(x^{3}-1\right)^{2}\right.$
$\left.+{ }^{6} C_{6}\left(x^{3}-1\right)^{3}\right]$
$=2\left[x^{6}+15 x^{7}-15 x^{4}+15 x^{8}-30 x^{5}+15 x^{2}+x^{9}-3 x^{6}\right.$
$\left.+3 x^{3}-1\right]$
Hence, the sum of coefficients of even powers of
$x=2[1-15+15+15-3-1]=24$