SATHEE: Binomial Theorem 1 Question 14

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

(d) 2

(2018 Main)

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Answer:

Correct Answer: 15. $\frac{1}{4} (n + 1)^{2} (2 n - 1)$

Solution:

Key Idea Use formula :

$\begin{aligned} = (a + b)^{n} + (a - b)^{n} \\ = 2 (^{n} C_{0} a^{n} +^{n} C_{2} a^{n - 2} b^{2} +^{n} C_{4} a^{n - 4} b^{4} + \dots) \end{aligned}$

We have, ${(x + \sqrt{x^{3} - 1})}^{5} + {(x - \sqrt{x^{3} - 1})}^{5}, x > 1$

$= 2 (^{5} C_{0} x^{5} +^{5} C_{2} x^{3} {(\sqrt{x^{3} - 1})}^{2} +^{5} C_{4} x {(\sqrt{x^{3} - 1})}^{4})$

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

$= 2 (x^{5} + 10 x^{6} - 10 x^{3} + 5 x^{7} - 10 x^{4} + 5 x)$

Sum of coefficients of all odd degree terms is

$2 (1 - 10 + 5 + 5) = 2$

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Binomial Theorem 1 Question 14

15. The sum of the coefficients of all odd degree terms in the expansion of x+x3−15+x−x3−15,(x>1) is

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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