If sum of the coefficients of x7 and x4 in the expansion of x2a−bx11 is zero, then

A
ab = 1
B
a = b
C
ab = –1
D
a + b = 0

Solution:

$\begin{array}{l} T_{r + 1} =^{11} C_{r} {(\frac{x^{2}}{a})}^{11 - r} {(- \frac{b}{x})}^{r} \\ =^{11} C_{r} (- 1)^{r} a^{r - 11} b^{r} x^{22 - 3 r} \end{array}$

For coefficient of x ⁴ and x⁷ , we put $22 - 3 r = 4$ and $22 - 3 r = 7$

$\Rightarrow r = 6$ and $r = 5$

We are given

$^{11} C_{6} (- 1)^{6} a^{- 5} b^{6} +^{11} C_{5} (- 1)^{5} a^{- 6} b^{5} = 0$

$\Rightarrow \frac{b^{6}}{a^{5}} = \frac{b^{5}}{a^{6}} \Rightarrow a b = 1$

If sum of the coefficients of x⁷ and x⁴in the expansion of ${(\frac{x^{2}}{a} - \frac{b}{x})}^{11}$ is zero, then

Solution:

Related content