SATHEE: Binomial Theorem 1 Question 7

Binomial Theorem 1 Question 7

8.

The sum of the coefficients of all even degree terms is $x$ in the expansion of

${(x + \sqrt{x^{3} - 1})}^{6} + {(x - \sqrt{x^{3} - 1})}^{6}, (x > 1)$ is equal to

(2019 Main, 8 April I)

(a) 29

(b) 32

(d) 24

Show Answer

Answer:

Correct Answer: 8. $(d)$

Solution:

Key Idea Use formula :

$(a + b)^{n} + (a - b)^{n} =$

$2 [^{n} C_{0} a^{n} +^{n} C_{2} a^{n - 2} b^{2} +^{n} C_{4} a^{n - 4} b^{4} + \dots \dots]$

Given expression is ${(x + \sqrt{x^{3} - 1})}^{6} + {(x - \sqrt{x^{3} - 1})}^{6}$

$= 2 [^{6} C_{0} x^{6} +^{6} C_{2} x^{4} {(\sqrt{x^{3} - 1})}^{2}$ $+^{6} C_{4} x^{2} {(\sqrt{x^{3} - 1})}^{4} +^{6} C_{6} {(\sqrt{x^{3} - 1})}^{6}]$

$∵ (a + b)^{n} + (a - b)^{n}$

$= 2 [^{n} C_{0} a^{n} +^{n} C_{2} a^{n - 2} b^{2} +^{n} C_{4} a^{n - 4} b^{4} + \dots]$

$= 2 [^{6} C_{0} x^{6} +^{6} C_{2} x^{4} (x^{3} - 1) +^{6} C_{4} x^{2} {(x^{3} - 1)}^{2} +^{6} C_{6} {(x^{3} - 1)}^{3}]$

The sum of the terms with even power of $x$

$\begin{aligned} = 2 [^{6} C_{0} x^{6} +^{6} C_{2} (- x^{4}) +^{6} C_{4} x^{8} +^{6} C_{4} x^{2} +^{6} C_{6} (- 1 - 3 x^{6})] \\ = 2 [^{6} C_{0} x^{6} -^{6} C_{2} x^{4} +^{6} C_{4} x^{8} +^{6} C_{4} x^{2} - 1 - 3 x^{6}] \end{aligned}$

Now, the required sum of the coefficients of even powers of $x$ in

$\begin{aligned} {(x + \sqrt{x^{3} - 1})}^{6} + {(x - \sqrt{x^{3} - 1})}^{6} \\ = 2 [^{6} C_{0} -^{6} C_{2} +^{6} C_{4} +^{6} C_{4} - 1 - 3] \\ = 2 [1 - 15 + 15 + 15 - 1 - 3] = 2 (15 - 3) = 24 \end{aligned}$

Binomial Theorem 1 Question 7

8.

Answer:

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Binomial Theorem 1 Question 7

8.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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