The absolute difference of the coefficients of $x^{10}$ and $x^{7}$ in the expansion of $\left(2 x^{2}+\frac{1}{2 x}\right)^{11}$ is equal to

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Binomial Theorem

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

Answer: (A)

The general term in the expansion is:

T_{r + 1} =^{11} C_{r} (2 x^{2})^{11 - r} (\frac{1}{2 x})^{r} =^{11} C_{r} 2^{11 - 2 r} x^{22 - 3 r}

For

x^{10}

22 - 3 r = 10 \Rightarrow r = 4

Coefficient:

^{11} C_{4} \cdot 2^{3} = 330 \cdot 8 = 2640

For

x^{7}

22 - 3 r = 7 \Rightarrow r = 5

Coefficient:

^{11} C_{5} \cdot 2^{1} = 462 \cdot 2 = 924

Absolute difference:

∣2640 - 924∣ = 1716 = 1 2^{3} - 12 = 1728 - 12 = 1716

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option