If the coefficient of x 7 in ( a x - 1 b x 2 ) 13 and the coefficient of x - 5 in ( a x + 1 b x 2 ) 13 are equal, then a 4 b 4 is equal to [2023]

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Solution

Q.
If the coefficient of $x^{7}$ in ${(a x - \frac{1}{b x^{2}})}^{13}$ and the coefficient of $x^{- 5}$ in ${(a x + \frac{1}{b x^{2}})}^{13}$ are equal, then $a^{4} b^{4}$ is equal to [2023]

1 11

2 44

3 22

4 33

Ans.
(3)

In the first expansion,
$T_{r + 1} = {}^{13}{C_{r}} {(a x)}^{13 - r} {(\frac{- 1}{b x^{2}})}^{r} = {}^{13}{C_{r}} a^{13 - r} {(\frac{- 1}{b})}^{r} x^{13 - 3 r}$

Now, $13 - 3 r = 7 \Rightarrow r = 2$

$∴$ $Coefficient of x^{7} = {}^{13}{C_{2}} \frac{a^{11}}{b^{2}}$

Also, in the second expansion,
$T_{r + 1} = {}^{13}{C_{r}} {(a x)}^{13 - r} {(\frac{1}{b x^{2}})}^{r} = {}^{13}{C_{r}} \frac{a^{13 - r}}{b^{r}} x^{13 - 3 r}$

As, $13 - 3 r = - 5 \Rightarrow r = 6$

$∴$ $Coefficient of x^{- 5} = {}^{13}{C_{6}} \frac{a^{7}}{b^{6}}$

Now, ${}^{13}{C_{2}} \frac{a^{11}}{b^{2}} = {}^{13}{C_{6}} \frac{a^{7}}{b^{6}}$ $\Rightarrow a^{4} b^{4} = \frac{{}^{13}{C_{6}}}{{}^{13}{C_{2}}} = 22$

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