The absolute difference of the coefficients of x 10 and x 7 in the expansion of ( 2 x 2 + 1 2 x ) 11 is equal to [2023]

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Solution

Q.
The absolute difference of the coefficients of $x^{10}$ and $x^{7}$ in the expansion of ${(2 x^{2} + \frac{1}{2 x})}^{11}$ is equal to [2023]

1 $10^{3} - 10$

2 $12^{3} - 12$

3 $13^{3} - 13$

4 $11^{3} - 11$

Ans.
(2)

We have, $T_{r + 1} = {}^{11}{C_{r}} {(2 x^{2})}^{11 - r} {(\frac{1}{2 x})}^{r} = {}^{11}{C_{r}} 2^{11 - 2 r} \cdot x^{22 - 3 r}$

For $x^{10}$ , $22 - 3 r = 10 \Rightarrow r = 4$

$∴$ $Coefficient of x^{10} = {}^{11}{C_{4}}$ $2^{11 - 8} = 330 \times 2^{3} = 2640$

Now, for $x^{7}$ , $22 - 3 r = 7 \Rightarrow r = 5$

$∴$ $Coefficient of x^{7} = {}^{11}{C_{5}} \cdot 2^{11 - 10} = 462 \times 2 = 924$

So, absolute difference of the coefficients of $x^{10}$ and $x^{7}$ is

$2640 - 924 = 1716 = 12^{3} - 12$

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