Question

If the coefficient of $x^7$ in $(a{x}^2+\frac{1}{bx}{)}^{11}$ is equal to the coefficient of $x^{-7}$ in $(ax-\frac{1}{b{x}^2}{)}^{11}$, then ab =

• A. 1
• B. $\frac{1}{2}$
• C. 2
• D. 3

Answer 1

The correct option is A

1

In the expansion of $(a{x}^2+\frac{1}{bx}{)}^{11}$, the general

term is

$T_{r+1}$ = ${}^{11}{C}_r$$(a{x}^2{)}^{11-r}$$(\frac{1}{bx}{)}^r$ = ${}^{11}{C}_r$$(a{)}^{11-r}$$\left(\frac{1}{b^r}\right)x^{22-3r}$

For $x^7$, we must have 22 - 3r = 7 ⇒ r = 5, and

the coefficient of $x^7$ = ${}^{11}{C}_5$.$a^{11-5}\frac{1}{5}$) = ${}^{11}{C}_5$ $\frac{a^6}{b^5}$

Similarly, in the expansion of $(ax-\frac{1}{b{x}^2}{)}^{11}$, 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, ${}^{11}{C}_5$ $\frac{a^6}{b^5}$ = ${}^{11}{C}_5$ $\frac{a^5}{b^6}$ ⇒ ab = 1.