Simplifying modulus expressions and an unknown expression? discrete math

Question

I have a few questions below that I need help with a) I don't really understand what that symbol means and how to solve it b) How do u simplify this without a calculator c) I got 2^-r = 0, iss this true?

Answer 1

For part (a):

1) $\prod_{k=2}^{n}\frac{k+1}{k}=\frac{3}{2}.\frac{4}{3}...\frac{n}{n-1}.\frac{n+1}{n}=\frac{n+1}{2}$

2) $\prod_{k=2}^{n}\frac{k-1}{k}=\frac{1}{2}.\frac{2}{3}...\frac{n-2}{n-1}.\frac{n-1}{n}=\frac{1}{n}$

$\prod_{k=2}^{n}(1-\frac{1}{k^2})= \prod_{k=2}^{n}\frac{k^2-1}{k^2}=\prod_{k=2}^{n}(\frac{k+1}{k})(\frac{k-1}{k})=\frac{n+1}{2n}$ (using the result from 1 and 2)

Part (b), see lab's answer.

Part(c):
Recall the sum of geometric series: $$\sum_{r=1}^{n}a^r=\frac{a-a^{n+1}}{1-a}$$
So, $$\sum_{r=1}^{\infty}(\frac{1}{2})^r=\frac{\frac{1}{2}-(\frac{1}{2})^\infty}{1-\frac{1}{2}}=\frac{\frac{1}{2}-0}{\frac{1}{2}}=1$$

Answer 2

$$3^3=27\equiv-1\pmod7\implies3^{1000}=3(3^3)^{333}\equiv3(-1)^{333}\equiv-3\equiv4$$

or $$3^6\equiv(-1)^2\pmod7\equiv1,1000\equiv4\pmod6\implies3^{1000}\equiv3^4\pmod7$$