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MISCELLANEOUS EXERCISE 4 (II) · Q115

Q.If the coefficient of x16x^{16} in the expansion of (x2+ax)10(x^2 + ax)^{10} is 33603360, find aa.

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tr+1=10Cr(x2)10−r(ax)r=10Crarx20−rt_{r+1}={}^{10}C_r(x^2)^{10-r}(ax)^r={}^{10}C_ra^rx^{20-r}. Setting 20−r=1620-r=16 gives r=4r=4. Coefficient =10C4a4=210a4={}^{10}C_4a^4=210a^4. Given 210a4=3360210a^4=3360, so a4=16a^4=16, …

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