If $lpha$ and $eta$ be the coefficients of
Question:

If $\alpha$ and $\beta$ be the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of

$\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}$, then:

 

  1. (1) $\alpha+\beta=60$

  2. (2) $\alpha+\beta=-30$

  3. (3) $\alpha-\beta=60$

  4. (4) $\alpha-\beta=-132$


Correct Option: , 4

Solution:

Using Binomial expansion

$(x+a)^{n}+(x-a)^{n}=2\left(T_{1}+T_{3}+T_{5}+T_{7} \cdots\right)$

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

$2\left[{ }^{6} C_{0} x^{5}+{ }^{6} C_{2} x^{4}\left(x^{2}-1\right)+{ }^{6} C_{4} x^{2}\left(x^{2}-1\right)^{2}\right.$

$\left.+{ }^{6} C_{6}\left(x^{2}-1\right)^{3}\right]$

$=2\left[x^{6}+15\left(x^{6}-x^{4}\right)+15 x^{2}\left(x^{4}-2 x^{2}+1\right)\right.$

$\left.+\left(-1+3 x^{2}-3 x^{4}+x^{6}\right)\right]$

$=2\left(32 x^{6}-48 x^{4}+18 x^{2}-1\right)$

$\alpha=-96$ and $\beta=36$

$\therefore \quad \alpha-\beta=-132$

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