Originally Posted by
e^(i*pi) I'll expand the first part for you and give guidance on the others but I'm not doing your work for you in full.
$\displaystyle
(1-p)^5 = 1 - 5p + \frac{5 \cdot 4}{2!}p^2 - \frac{5 \cdot 4 \cdot 3}{3!}p^3 + \frac{5 \cdot 4 \cdot 3 \cdot 2}{4!}p^4 - \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5!}p^5$
Any further terms will have multiplication by 0 so there are no more terms. I'll leave simplifying that to you.
What the next question is asking is to substitute $\displaystyle (x+x^2)$ for p in the above expansion. You can either do this as above or substitute $\displaystyle p = x(1+x)$, if you use the latter remember that the whole expression is squared.
For the third part it is asking you to solve $\displaystyle 1-q = 0.9899$ and $\displaystyle q = x^2+x$ where q is a constant.