Q.

The last term in the binomial expansion of2−12n   is  1(3)913log⁡3then the 5th term from the beginning is

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a

10C6

b

2 10C4

c

12 10C4

d

−10C6

answer is B.

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Detailed Solution

nCn−12n=3−1−2/3log3⁡8=3−5log3⁡2=125⇒ n=10 ∴  5th term from the beginning =10C4(2)10−4−124=2 10C4
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