How would I write this as an integer or a fraction in lowest terms?
$(1-\frac12)(1+\frac 12)(1-\frac13)(1+\frac13)(1-\frac14)(1+\frac14).....(1-\frac1{99})(1+\frac1{99})$
I really need to understand where to start and the process if anyone can help me.
On
Rewrite it instead as
$$\color{Green}{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdots\left(1-\frac{1}{98}\right)\left(1-\frac{1}{99}\right)}\times \color{Blue}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{98}\right)\left(1+\frac{1}{99}\right)}$$
Evaluate left and right parts separately. Hint: Look at how numerator and denominator cancel.
For example,
$$\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)=\frac{\color{DarkOrange}1}{\color{Red}2}\cdot\frac{\color{Red}2}{\color{Green}3}\cdot\frac{\color{Green}3}{\color{Blue}4}\cdot\frac{\color{Blue}4}{\color{DarkOrange}5}=\frac{1}{5}.$$
Hint: telescopy, note how the adjacent like-colored terms all cancel out of the products below
$$\rm \left(1-\frac{1}2\right)\left(1-\frac{1}3\right)\cdots \left(1-\frac{1}n\right)\ =\ \frac{1}{\color{red}2} \frac{\color{red}2}{\color{green}3} \frac{\color{green}3}{\color{blue}4} \frac{\color{blue} 4}{}\: \cdots\: \frac{}{\color{brown}{n-1}}\frac{\color{brown}{n-1}}n\ =\ \frac{1}n $$
$$\rm \left(1+\frac{1}2\right)\left(1+\frac{1}3\right)\cdots \left(1+\frac{1}n\right)\ =\ \frac{\color{red}3}2 \frac{\color{green}4}{\color{red}3} \frac{\color{blue}5}{\color{green}4}\frac{}{\color{blue}5}\: \cdots\: \frac{\color{brown} n}{}\frac{n+1}{\color{brown}n}\ =\ \frac{n+1}2$$
Hence $$\rm \prod_{k\:=\:2}^n\: \left( 1-\frac{1}{k^2}\right)\ =\ \frac{n+1}{2\:n} $$
Follow the above link for many more examples of telescopic proofs (additive and multiplicative).