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Exercise: General Term · Q18

Q.If the coefficients of x2x^2 and x3x^3 in the expansion of (3+ax)9(3+ax)^9 are equal, find the value of aa.

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The general term of (3+ax)9(3+ax)^9 is Tr+1=9Cr 39−r(ax)r=9Cr 39−rar xrT_{r+1}={}^{9}C_r\,3^{9-r}(ax)^r={}^{9}C_r\,3^{9-r}a^r\,x^r. The coefficient of x2x^2 (r=2r=2) is 9C2 37a2=36×2187 a2=78732a2^{9}C_2\,3^7a^2=36\times2187\,a^2=78732a^2. The coefficient of x3x^3 (r=3r=3) is 9C3 36a3=84×729 a3=61236a3^{9}C_3\,3^6a^3=84\times729\,a^3=61236a^3. Setting them equal: 78732a2=61236a378732a^2=61236a^3. Since a≠0a\ne0, divide both sides by a2a^2: $78 …

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