Coefficient of x Squared in Binomial Expansion

To find the coefficient of $x^2$ in the binomial expansion of $\left(2x^2 + \frac{1}{x}\right)^{10}$, we start by writing the general term of the expansion.

The general term $T_{r+1}$ in the expansion of $(a + b)^n$ is given by: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$

For the given expression $\left(2x^2 + \frac{1}{x}\right)^{10}$, where $a = 2x^2$, $b = x^{-1}$, and $n = 10$, the general term is: $T_{r+1} = \binom{10}{r} \left(2x^2\right)^{10-r} \left(\frac{1}{x}\right)^r$

Simplifying the powers of $x$: $T_{r+1} = \binom{10}{r} 2^{10-r} \cdot x^{2(10-r)} \cdot x^{-r}$ $T_{r+1} = \binom{10}{r} 2^{10-r} \cdot x^{20-3r}$

To find the coefficient of $x^2$, we set the exponent of $x$ equal to $2$: $20 - 3r = 2$ $3r = 18 \implies r = 6$

Since $r = 6$ is an integer such that $0 \le r \le 10$, the term containing $x^2$ exists. Substituting $r = 6$ into the coefficient part of the general term gives: $\text{Coefficient of } x^2 = \binom{10}{6} 2^{10-6}$

Using the property $\binom{n}{r} = \binom{n}{n-r}$, we have $\binom{10}{6} = \binom{10}{4}$: $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Also, $2^{10-6} = 2^4 = 16$.

Thus, the coefficient of $x^2$ is: $\text{Coefficient} = 210 \times 16 = 3360$

Hence, the correct option is B.