Binomial Theorem Ques 40
The sum of the coefficients of all even degree terms is $x$ 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
(2019 Main, 8 April I)
(a) 29
(b) 32
(c) 26
(d) 24
Show Answer
Answer:
Correct Answer: 40.$(d)$
Solution:
Key Idea Use formula :
$(a+b)^{n}+(a-b)^{n}=$
$ 2\left[{ }^{n} C_{0} a^{n}+{ }^{n} C_{2} a^{n-2} b^{2}+{ }^{n} C_{4} a^{n-4} b^{4}+\ldots \ldots\right] $
Given expression is $\left(x+\sqrt{x^{3}-1}\right)^{6}+\left(x-\sqrt{x^{3}-1}\right)^{6}$
$ =2[{ }^{6} C_{0} x^{6}+{ }^{6} C_{2} x^{4}\left(\sqrt{x^{3}-1}\right)^{2} $ $ +{ }^{6} C_{4} x^{2}\left(\sqrt{x^{3}-1}\right)^{4}+{ }^{6} C_{6}\left(\sqrt{x^{3}-1}\right)^{6}]$
$ \because(a+b)^{n}+(a-b)^{n} $
$ =2[{ }^{n} C_{0} a^{n}+{ }^{n} C_{2} a^{n-2} b^{2}+{ }^{n} C_{4} a^{n-4} b^{4}+\ldots]$
$ =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}+{ }^{6} C_{6}\left(x^{3}-1\right)^{3}\right] $
The sum of the terms with even power of $x$
$ \begin{aligned} & =2\left[{ }^{6} C_{0} x^{6}+{ }^{6} C_{2}\left(-x^{4}\right)+{ }^{6} C_{4} x^{8}+{ }^{6} C_{4} x^{2}+{ }^{6} C_{6}\left(-1-3 x^{6}\right)\right] \\ & =2\left[{ }^{6} C_{0} x^{6}-{ }^{6} C_{2} x^{4}+{ }^{6} C_{4} x^{8}+{ }^{6} C_{4} x^{2}-1-3 x^{6}\right] \end{aligned} $
Now, the required sum of the coefficients of even powers of $x$ in
$ \begin{aligned} & \left(x+\sqrt{x^{3}-1}\right)^{6}+\left(x-\sqrt{x^{3}-1}\right)^{6} \\ & \quad=2\left[{ }^{6} C_{0}-{ }^{6} C_{2}+{ }^{6} C_{4}+{ }^{6} C_{4}-1-3\right] \\ & \quad=2[1-15+15+15-1-3]=2(15-3)=24 \end{aligned} $
BETA
