The sum of co-efficients of all even degree terms in x in the expansion of (x+√x3−1)6+(x−√x3−1)6,(x>1) is equal to:
A
24
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B
26
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C
29
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D
32
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Solution
The correct option is A24 If we add terms of both the expansions, even numbered terms will cancel each other and we get- 2[6C0x6+6C2x4(√x3−1)2+6C4x2(√x3−1)4+6C6(√x3−1)6]=2[6C0x6+6C2x4(x3−1)+6C4x2(x3−1)2+6C6(x3−1)3]=2[x6+6C2x7−6C2x4+6C4x8+6C4x2−26C4x5+x9−1−3x6+3x3]
Hence, sum of coefficients of even powers of x =2[1−6C2+6C4+6C4−1−3]=2[1−15+15+15−4]=24