Find the constant term in the binomial expansion \((2x^{2} + \frac{1}{x})^{9}\)
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Correct Answer: Option D
Explanation:
Let the power of \(2x^{2}\) be t and the power of \(\frac{1}{x} \equiv x^{-1}\) = 9 - t.
\((2x^{2})^{t}(x^{-1})^{9 - t} = x^{0}\)
Dealing with x alone, we have
\((x^{2t})(x^{-9 + t}) = x^{0} \implies 2t - 9 + t = 0\)
\(3t - 9 = 0 \therefore t = 3\)
The binomial expansion is then,
\(^{9}C_{3} (2x^{2})^{3}(x^{-1})^{6} = \frac{9!}{(9-3)! 3!} \times 2^{3}\)
= 84 x 8
= 672
Let the power of \(2x^{2}\) be t and the power of \(\frac{1}{x} \equiv x^{-1}\) = 9 - t.
\((2x^{2})^{t}(x^{-1})^{9 - t} = x^{0}\)
Dealing with x alone, we have
\((x^{2t})(x^{-9 + t}) = x^{0} \implies 2t - 9 + t = 0\)
\(3t - 9 = 0 \therefore t = 3\)
The binomial expansion is then,
\(^{9}C_{3} (2x^{2})^{3}(x^{-1})^{6} = \frac{9!}{(9-3)! 3!} \times 2^{3}\)
= 84 x 8
= 672