First slide

Binomial theorem for positive integral Index

Question

If the coefficient of x⁷ in the expansion of ${(a x^{2} + \frac{1}{b x})}^{11}$ and the coefficient of x^{– 7}is the expansion of ${(a x - \frac{1}{b x^{2}})}^{11}$ are equal, then

Moderate

A

ab = 1
B

ab = 11
C

ab = 5
D

ab = 6

Solution

$T_{r + 1},$ the ${(r + 1)}^{th}$ term in the expansion ${(a x^{2} + \frac{1}{b x})}^{11}$ is
$T_{r + 1} =^{11} C_{r} {(a x^{2})}^{11 - r} {(\frac{1}{b x})}^{r} =^{11} C_{r} \frac{a^{11 - r}}{b^{r}} x^{22 - 3 r}$
For the coefficient of $x^{7},$ set $22 - 3 r = 7 \Rightarrow r = 5$
$∴$ Coefficient of x⁷ in ${(a x^{2} + \frac{1}{b x})}^{11}$ is $^{11} C_{5} \frac{a^{6}}{b^{5}}$
Next, $t_{r} + 1$ , the ${(r + 1)}^{t h}$ term in the expansion of ${(a x - \frac{1}{b x^{2}})}^{11}$ is
$t_{r + 1} =^{11} C_{r} (a x)^{11 - r} {(- \frac{1}{b x^{2}})}^{r}$
$=^{11} C_{r} \frac{a^{11 - r}}{b^{r}} (- 1)^{r} x^{11 - 3 r}$
For the coefficient of $x^{- 7},$ set $11 - 3 r = - 7 \Rightarrow r = 6$
$∴$ Coefficient of $x^{- 7}$ in the expansion of $=^{11} C_{6} \frac{a^{5}}{b^{6}}$
We are given $^{11} C_{5} \frac{a^{6}}{b^{5}} =^{11} C_{6} \frac{a^{5}}{b^{6}} \Rightarrow a b = 1$

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