It's 60.
Using the binomial theorem,the fifth term is (6)(5)(4)(3)(2x^2)^2(1/x)^4 / 4!
=(6)(5)(4)(3)(4x^4)(1/x^4) /4!
=(6)(5)(4)(3) (4) /4!
= (6)(5)(4)(3) (4)/ (4)(3)(2)
=(6)(5)(4) /(2) = (6)(5)(2) = 30 x 2 = 60
+129820 The constant willl occur at the 5th term in the binomial expansion of this =
C(6,4) * (2x^2)^2 * (1/x)^4 =
15 (4x^4) (1/x^4) =
15* 4 =
60
+26383 how do i find the constant term in the expansion of (2x^2-1/x)^6
\(\begin{array}{|rcll|} \hline && \left( 2x^2-\frac{1}{x} \right)^6 \\\\ &=& \left( \frac{2x^3-1}{x} \right)^6 \\\\ &=& \frac{(2x^3-1)^6}{x^6} \\\\ &=& \frac{1}{x^6} \cdot (2x^3-1)^6 \\ \hline \end{array} \)
The constant term is at the 5th term in the binomial expansion:
\(\begin{array}{|rcll|} \hline && \frac{1}{x^6} \cdot \binom64 \cdot (2x^3)^2 \quad & | \quad \binom64 = \binom{6}{6-4} = \binom62\\ &=& \frac{1}{x^6} \cdot \binom62 \cdot (2x^3)^2 \\ &=& \frac{1}{x^6} \cdot \binom62 \cdot 4x^6 \\ &=& 4 \cdot \binom62 \\ &=& 4 \cdot \frac62\cdot \frac51 \\ &=& 4 \cdot 3\cdot 5 \\ &=& 60 \\ \hline \end{array} \)
