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Suppose $a \in R .$ If the coefficient of $x^{5}$ in the expansion of ${(a x + \frac{1}{x^{3}})}^{17}$ is $680,$ then $a$ is equal to

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Correct option is B

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

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Ready to Test Your Skills?

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Suppose $a \in R .$ If the coefficient of $x^{5}$ in the expansion of ${(a x + \frac{1}{x^{3}})}^{17}$ is $680,$ then $a$ is equal to