First slide

Binomial theorem for positive integral Index

Question

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 equa to

Moderate

A

29
B

32
C

26
D

24

Solution

Given expression is ${(x + \sqrt{x^{3} - 1})}^{6} + {(x - \sqrt{x^{3} - 1})}^{6}$
$= 2 [^{6} C_{0} x^{6} +^{6} C_{2} x^{4} {(\sqrt{x^{3} - 1})}^{2} +^{6} C_{4} x^{2} {(\sqrt{x^{3} - 1})}^{4} +^{6} C_{6} {(\sqrt{x^{3} - 1})}^{6}]$
$\{∵ (a + b)^{n} + (a - b)^{n}$
$\begin{array}{l} = 2 [^{n} C_{0} a^{n} +^{n} C_{2} a^{n - 2} b^{2} +^{n} C_{4} a^{n - 4} b^{4} + \dots]\} \\ ^{6} C_{0} x^{6} +^{6} C_{2} x^{4} (x^{3} - 1) +^{6} C_{4} x^{2} {(x^{3} - 1)}^{2} +^{6} C_{6} {(x^{3} - 1)}^{3}] \end{array}$
The sum of the terms with even power of x
$\begin{array}{l} = 2 [^{6} C_{0} x^{6} +^{6} C_{2} (- x^{4}) +^{6} C_{4} x^{8} +^{6} C_{4} x^{2} +^{6} C_{6} (- 1 - 3 x^{6})] \\ = 2 [^{6} C_{0} x^{6} -^{6} C_{2} x^{4} +^{6} C_{4} x^{8} +^{6} C_{4} x^{2} - 1 - 3 x^{6}] \end{array}$
Now, the required sum of the coefficients of even powers of x in ${(x + \sqrt{x^{3} - 1})}^{6} + {(x - \sqrt{x^{3} - 1})}^{6}$
$\begin{array}{l} = 2 [^{6} C_{0} -^{6} C_{2} +^{6} C_{4} +^{6} C_{4} - 1 - 3] \\ = 2 [1 - 15 + 15 + 15 - 1 - 3] = 2 (15 - 3) = 24 \end{array}$

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