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Miscellaneous · Q27

Q.Find the term independent of xx in the expansion of (32x2−13x)9\left(\dfrac{3}{2}x^2-\dfrac{1}{3x}\right)^9.

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General term: Tr+1=9Cr(32x2)9−r(−13x)r=9Cr(−1)r39−r29−r⋅13r x2(9−r)−r=9Cr(−1)r39−2r29−r x18−3rT_{r+1}={}^{9}C_r\left(\dfrac{3}{2}x^2\right)^{9-r}\left(-\dfrac{1}{3x}\right)^r={}^{9}C_r(-1)^r\dfrac{3^{9-r}}{2^{9-r}}\cdot\dfrac{1}{3^r}\,x^{2(9-r)-r}={}^{9}C_r(-1)^r\dfrac{3^{9-2r}}{2^{9-r}}\,x^{18-3r}. Setting 18−3r=018-3r=0 gives r=6r=6. Substituting: $T_7={}^{9}C_6(-1)^6\dfrac{3^{9-12}}{2^{9-6}}={}^{9}C_6\cdot\dfrac{3^{-3}}{2^3}=84\times\dfrac{1}{27\times8}=\d …

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