Find the independant of x in the expansion of [x-(2/x)]^4
Thanks, Rob
Note: $\displaystyle \left(x-\frac{2}{x}\right)^4$ can be written as $\displaystyle (x-2(x^{-1}))^4$
Just looking at the x terms (ignoring coefficients):
$\displaystyle (x^4)(x^0) + (x^3)(x^{-1}) + (x^2)(x^{-2}) + (x^1)(x^{-3}) + (x^0)(x^{-4})$
The third term cancels out.
Hello, Rob!
Don't argue about it . . . just expand it!
Find the constant term in the expansion of .$\displaystyle \left(x-\frac{2}{x}\right)^4$
$\displaystyle \left(x-\frac{2}{x}\right)^4 \;\;=\;\;(x)^4 - \:4(x)^3\!\left(\frac{2}{x}\right) + \:6(x)^2\!\left(\frac{2}{x}\right)^2 - \:4(x)^1\!\left(\frac{2}{x}\right)^3 + \left(\frac{2}{x}\right)^4$
. . . . . . . $\displaystyle = \;\;x^4 \:- \;4x^3\!\cdot\!\frac{2}{x} \:+ \;6x^2\!\cdot\!\frac{4}{x^2} \:- \;4x\!\cdot\!\frac{8}{x^3} \:+ \;\frac{16}{x^4}$
. . . . . . . $\displaystyle = \;\;x^4 \:-\: 8x^2 \:+\: \underbrace{{\color{blue}24}}_{\uparrow} \:-\: \frac{32}{x^2} \:+\: \frac{16}{x^4}$