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Exercise: General Term · Q16

Q.Find the coefficient of x7x^7 in the expansion of (x2+3x)11\left(x^2+\dfrac{3}{x}\right)^{11}.

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The general term of (x2+3x)11\left(x^2+\dfrac{3}{x}\right)^{11} is Tr+1=11Cr (x2)11−r(3x)r=11Cr 3r x22−2r−r=11Cr 3r x22−3rT_{r+1}={}^{11}C_r\,(x^2)^{11-r}\left(\dfrac{3}{x}\right)^r={}^{11}C_r\,3^r\,x^{22-2r-r}={}^{11}C_r\,3^r\,x^{22-3r}. Setting 22−3r=722-3r=7 gives 3r=153r=15, so r=5r=5. The coefficient is 11C5⋅35^{11}C_5\cdot3^5. Since $^{11}C_5=\dfrac{ …

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