Question:

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

$13^3-13$

$11^3-11$

$10^3-10$

$12^3-12$

Step-by-Step Solution

Key Concept: $T_{r+1}=\binom{11}{r}(2x^2)^{11-r}(\frac{1}{2x})^r=\binom{11}{r}2^{11-2r}x^{22-3r}$. For $x^7$: $r=5$; for $x^{10}$: $r=4$.

$|924-2640|=1716=12^3-12$.

Correct Answer: 4