Discussion

The constant term in the expansion of \(\left(2x^{2} - \frac{1}{x^{2}}\right)^{6}\) is

Answer: A

The constant term is −160.

The general term of \(\left(2x^{2} - \frac{1}{x^{2}}\right)^{6}\) is

\[ \begin{aligned} T_{r+1} &= {}^{6}C_{r}(2x^{2})^{6-r} \\ &\quad \times \left(-\frac{1}{x^{2}}\right)^{r} \end{aligned} \]

with power of x \(12 - 4r\). Setting it to 0 gives \(r = 3\):

\[ \begin{aligned} {}^{6}C_{3} \cdot 2^{3} \cdot (-1)^{3} &= 20 \times 8 \times (-1) \\ &= -160 \end{aligned} \]

160 drops the sign from \((-1)^{3}\), and ±180 use the wrong binomial coefficient or power of 2.

No comment is present. Be the first to comment.

Loading…