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If the coefficient of ${x^7}$ in ${\left( {a{x^2} + \frac{1}{{bx}}} \right)^{11}}$ is equal to the coefficient of ${x^{ - 7}}$ in ${\left( {ax - \frac{1}{{b{x^2}}}} \right)^{11}}$, then $ab =$
$1$
$1\over2$
$2$
$3$
Solution
(a) In the expansion of ${\left( {a{x^2} + \frac{1}{{bx}}} \right)^{11}}$, the general term is
${T_{r + 1}} = {\,^{11}}{C_r}{(a{x^2})^{11 – r}}{\left( {\frac{1}{{bx}}} \right)^r} = {\,^{\,11}}{C_r}{a^{11 – r}}\frac{1}{{{b^r}}}{x^{22 – 3r}}$
For $x^7$, we must have $22 -3r = 7$ ==> $r = 5$, and the coefficient of $x^7$ =$^{11}{C_5}.{a^{11 – 5}}\frac{1}{{{b^5}}} = {\,^{11}}{C_5}\frac{{{a^6}}}{{{b^5}}}$
Similarly, in the expansion of ${\left( {ax – \frac{1}{{b{x^2}}}} \right)^{11}},$ the general term is ${T_{r + 1}} = {\,^{11}}{C_r}{( – 1)^r}\frac{{{a^{11 – r}}}}{{{b^r}}}.{x^{11 – 3r}}$
For $x^{-7}$ we must have, $11 -3r = -7$ ==> $r = 6$, and the coefficient of ${x^{ – 7}}$ is $^{11}{C_6}\frac{{{a^5}}}{{{b^6}}} = {\,^{11}}{C_5}\frac{{{a^5}}}{{{b^6}}}$.
As given, $^{11}{C_5}\frac{{{a^6}}}{{{b^5}}} = {\,^{11}}{C_5}\frac{{{a^5}}}{{{b^6}}} \Rightarrow ab = 1$.
