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MISCELLANEOUS EXERCISE 4 (II) · Q119

Q.Show that there is no term containing x6x^6 in the expansion of (x2−3x)11\left(x^2-\dfrac{3}{x}\right)^{11}.

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tr+1=11Cr(x2)11−r(−3x)r=11Cr(−3)rx22−3rt_{r+1}={}^{11}C_r(x^2)^{11-r}\left(-\dfrac3x\right)^r={}^{11}C_r(-3)^rx^{22-3r}. For a term in x6x^6 we would need 22−3r=622-3r=6, i.e. 3r=163r=16, r=163r=\dfrac{16}3, which is not a whole number. Since rr must be an integer with 0≤r≤110\le r\le11, no such term exists. …

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