The coefficient of \(x^2\) in the expansion of \(\left(2 x^2+\frac{1}{x}\right)^{10}, x \neq 0\), is: [JEE Main 2026, 5 …

\(\left(2 x^2+\frac{1}{x}\right)^{10}, x \neq 0 \)
\( T_{r+1}={ }^{10} C_r\left(2 x^2\right)^{10-r}\left(\frac{1}{x}\right)^r \)
\( T_{r+1}={ }^{10} C_r(2)^{10-r} \cdot(x)^{20-3 r}\)
We need to find coefficient of \(x^2\)

\(\therefore 20-3 r=2 \)
\( 3 r=18 \Rightarrow r=6 \)
\( \therefore\) Coefficient of \( x^2={ }^{10} C_6(2)^{10-6} \)
\( ={ }^{10} C_6(2)^4=210 \times 16=3360\)