Question

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

• A. 1
• B. ½
• C. 2
• D. 3

Answer 1

Answer: (A)

1

Answer 2

For coefficient of $x^{7}$ in $\left( ax^{2} + \frac{1}{bx} \right)^{11}$; n = 11, $\alpha = 2,\beta = 1$, $m = 7$

$r = \frac{11.2 - 7}{2 + 1} = \frac{15}{3} = 5$

Coefficient of $x^{7}$ in $T_{6} =^{11} ⥂ C_{5}a^{6}.\frac{1}{b^{5}}$ ......(i)

and for coefficient of $x^{- 7}$ in $\left( ax - \frac{1}{bx^{2}} \right)^{11}$; $n = 11,\alpha = 1,\beta = 2$, $m = - 7$; $r = \frac{11.1 + 7}{3} = 6$

Coefficient of $x^{- 7}$ in $T_{7} =^{11} ⥂ C_{6}.a^{5}.\frac{1}{b^{6}}$ .....(ii)

From equation (i) and (ii), we get $ab = 1$