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MISCELLANEOUS EXERCISE 4 (II) · Q111

Q.Find the constant term in the expansion of (2x2−1x)12\left(2x^2-\dfrac{1}{x}\right)^{12}.

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tr+1=12Cr(2x2)12−r(−1x)r=12Cr212−r(−1)rx24−3rt_{r+1}={}^{12}C_r(2x^2)^{12-r}\left(-\dfrac1x\right)^r={}^{12}C_r2^{12-r}(-1)^rx^{24-3r}. Setting 24−3r=024-3r=0 gives r=8r=8. Constant $={}^{12}C_8(2)^4(-1 …

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