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MISCELLANEOUS EXERCISE 4 (II) · Q110

Q.Find the constant term in the expansion of (4x23+32x)9\left(\dfrac{4x^2}{3}+\dfrac{3}{2x}\right)^9.

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tr+1=9Cr(4x23)9−r(32x)r=9Cr(43)9−r(32)rx18−3rt_{r+1}={}^9C_r\left(\dfrac{4x^2}3\right)^{9-r}\left(\dfrac3{2x}\right)^r={}^9C_r\left(\dfrac43\right)^{9-r}\left(\dfrac32\right)^rx^{18-3r}. Setting 18−3r=018-3r=0 gives r=6r=6. Constant $={}^9C_6\left(\dfrac43\right)^3\left(\dfrac32\right)^6=84\times\dfra …

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