Solution :
We know that the (r + 1)th term or general term in the expansion of \((x + a)^n\) is given by
\(T_{r + 1}\) = \(^{n}C_r x^{n – r} a^r\)
In the expansion of \((2x^2 + {1\over x})^{12}\), we have
\(T_{10}\) = \(T_{9 + 1}\) = \(^{12}C_9 (2x^2)^{12 – 9} ({1\over x})^9\)
\(\implies\) \(T_{10}\) = \(^{12}C_9 (2x^2)^3 {1\over x^9}\)
\(\implies\) \(T_{10}\) = \(^{12}C_9\) \(8\over x^3\) = \(1760\over x^3\)
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