# Thread: Binomial expansion

1. ## Binomial expansion

Hello, i am unsure how to do this question... Find the value of the term in the expansion of (1-1/x)^8 which is independent of x. How do i do this? I have no problem with normal expansions but how do you do one with a fraction in?

Thankyou,
Kris.

2. I assume independent of x means it has no x term and is just a constant.

Then it's a matter of observation. It's 1. Becasue of the 1 in $(1-\frac{1}{x})^{8}$

If you have $(p+q)^{8}$, just sub in -1/x where the q goes in the expansion. Fractions are no different than an integer.

$1^{8}+8(\frac{-1}{x})+28(\frac{-1}{x})^{2}+...........+(\frac{-1}{x})^{8}$

3. Ok thankyou, yes i am not really sure what is meant by 'independent of x' as the expresion is (x-1/x)^8 is this just the constant in the expansion?

Thankyou for your help,
Kris

4. I suspect there is a typo in your original post: you probably want the constant term in the expansion of $\left(\color{red}x\color{black}-\frac{1}{x}\right)^8$. This would be the term $^8\mathrm{C}_4\,x^4\,\left(-\frac{1}{x}\right)^4$.