Suppose a ∈ R. If the coefficient of x5 in the expansion of ax+1x317is 680,then a is equal to

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±2

±1

±1/2

±1/3

answer is B.

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Detailed Solution

Tr+1=17Cr(ax)17−r1x3r=17Cra17−rx17−4rFor the coefficient of x5, we set 17−4r=5 or r=3.∴ 17C3a14=680⇒a14=680×3!14!17!=1⇒a=±1.

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