Binomial Theorem 1 Question 14
15. The sum of the coefficients 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) 0
(c) 1
(d) 2
(2018 Main)
Show Answer
Answer:
Correct Answer: 15. $\frac{1}{4}(n+1)^{2}(2 n-1)$
Solution:
Key Idea Use formula :
$$ \begin{aligned} & =(a+b)^{n}+(a-b)^{n} \\ & =2\left({ }^{n} C _0 a^{n}+{ }^{n} C _2 a^{n-2} b^{2}+{ }^{n} C _4 a^{n-4} b^{4}+\ldots\right) \end{aligned} $$
We have, $\left(x+\sqrt{x^{3}-1}\right)^{5}+\left(x-\sqrt{x^{3}-1}\right)^{5}, x>1$
$=2\left({ }^{5} C _0 x^{5}+{ }^{5} C _2 x^{3}\left(\sqrt{x^{3}-1}\right)^{2}+{ }^{5} C _4 x\left(\sqrt{x^{3}-1}\right)^{4}\right)$
$=2\left(x^{5}+10 x^{3}\left(x^{3}-1\right)+5 x\left(x^{3}-1\right)^{2}\right)$
$=2\left(x^{5}+10 x^{6}-10 x^{3}+5 x^{7}-10 x^{4}+5 x\right)$
Sum of coefficients of all odd degree terms is
$$ 2(1-10+5+5)=2 $$
