Question

If the coefficients of $x^7$ in ${(x^2+\frac{1}{bx})}^{11}$ and $x^{-7}$ in ${(x-\frac{1}{b{x}^2})}^{11},b\ne 0,$ are equal, then the value of $b$ is equal to

• A. $1$
• B. $-2$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $1$

The general term in ${(x^2+\frac{1}{bx})}^{11}$ is
$T_{r+1}={}^{11}{C}_r\cdot (x^2{)}^{11-r}\cdot {\left(\frac{1}{bx}\right)}^r$
$\Rightarrow T_{r+1}=\frac{{}^{11}{C}_r}{b^r}\cdot x^{22-3r}$
Coefficient of $x^7=\frac{{}^{11}{C}_5}{b^5}$

Similarly, the general term in${(x-\frac{1}{b{x}^2})}^{11}$ is
$T_{r+1}={}^{11}{C}_r\cdot (x{)}^{11-r}\cdot {(-\frac{1}{b{x}^2})}^r$
$\Rightarrow T_{r+1}=(-1{)}^r\cdot \frac{{}^{11}{C}_r}{b^r}\cdot x^{11-3r}$
Coefficient of $x^{-7}=\frac{{}^{11}{C}_6}{b^6}$
Now, according to the question, we have
$\frac{{}^{11}{C}_5}{b^5}=\frac{{}^{11}{C}_6}{b^6}$
$\Rightarrow b=\frac{{}^{11}{C}_6}{{}^{11}{C}_6}$
$\therefore b=1$