Find the constant term in the binomial expansion ((2x^{2} + frac{1}{x})^{9})

Find the constant term in the binomial expansion ((2x^{2} + frac{1}{x})^{9})

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