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(B) The general term of $\left(a x+\frac{b}{x^2}\right)^{11}$ is $T_{r+1} = {}^{11}C_r (ax)^{11-r} \left(\frac{b}{x^2}\right)^r = {}^{11}C_r a^{11-r}

Vedclass · Accepted Answer

$\frac{1}{a}\left[	ext{coefficient of } x^6 	ext{ in } \left(a x^2+\frac{b}{x}ight)^{12}ight]$

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(B) The general term of $\left(a x+\frac{b}{x^2}ight)^{11}$ is $T_{r+1} = {}^{11}C_r (ax)^{11-r} \left(\frac{b}{x^2}ight)^r = {}^{11}C_r a^{11-r} b^r x^{11-3r}$.For the coefficient of $x^{-7}$,set $11-3r = -7$,which gives $3r = 18$,so $r = 6$.Thus,$L = {}^{11}C_6 a^5 b^6$.Similarly,the general term of $\left(b x^2+\frac{a}{x}ight)^{11}$ is $T_{r+1} = {}^{11}C_r (bx^2)^{11-r} \left(\frac{a}{x}ight)^r = {}^{11}C_r b^{11-r} a^r x^{22-3r}$.For the coefficient of $x^7$,set $22-3r = 7$,which gives $3r = 15$,so $r = 5$.Thus,$M = {}^{11}C_5 b^6 a^5 = {}^{11}C_6 a^5 b^6$ (since ${}^{11}C_5 = {}^{11}C_6$).Therefore,$L+M = 2 	imes {}^{11}C_6 a^5 b^6$.Now,consider the general term of $\left(ax^2+\frac{b}{x}ight)^{12}$,which is $T_{r+1} = {}^{12}C_r (ax^2)^{12-r} \left(\frac{b}{x}ight)^r = {}^{12}C_r a^{12-r} b^r x^{24-3r}$.For the coefficient of $x^6$,set $24-3r = 6$,which gives $3r = 18$,so $r = 6$.The coefficient is ${}^{12}C_6 a^6 b^6$.Note that ${}^{12}C_6 = \frac{12}{6} 	imes {}^{11}C_5 = 2 	imes {}^{11}C_6$.So,the coefficient of $x^6$ is $2 	imes {}^{11}C_6 a^6 b^6 = a(2 	imes {}^{11}C_6 a^5 b^6) = a(L+M)$.Thus,$L+M = \frac{1}{a} \left[	ext{coefficient of } x^6 	ext{ in } \left(ax^2+\frac{b}{x}ight)^{12}ight]$.

If $L$ and $M$ are respectively the coefficient of $x^{-7}$ in $\left(a x+\frac{b}{x^2}\right)^{11}$ and the coefficient of $x^7$ in $\left(b x^2+\frac{a}{x}\right)^{11}$,then $L+M=$

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