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

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.