Question

I NEED AN ANSWER ASAP PLEASE INCLUDES STEP BY STEP Express as a common fraction in lowest term: f=%281-%5Cfrac%7B1%7D%7B2%7D%29%281-%5Cfrac%7B1%7D%7B3%7D%29%281-%5Cfrac%7B1%7D%7B4%7D%29...%281-%5Cfrac%7B1%7D%7B100%7D%29" id="TexFormula1" title="(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{100})" alt="(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{100})" align="absmiddle" class="latex-formula">

Answer 1

Answer:

1/100

Step-by-step explanation:

Answer 2

Answer:

1/100

Step-by-step explanation:

First, note that another way to write this is:

$(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4} ) ... (1-\frac{1}{100} )$

$=(\frac{1}{2})(\frac{2}{3})(\frac{3}{4} )... (\frac{99}{100} )$

For each subsequent term in the numerator (starting at 1), it increases by 1.

For each subsequent term in the denominator (starting at 2), it increases by 1.

Thus, notice across the numerator; we have:

$(1)(2)(3)(4) ... (99)$ or $99!$

Across the denominator we have

$(2)(3)(4) ... (100) = (1)(2)(3)(4) ... (100)=100!$

So, all together:

$=\frac{99!}{100!} =\frac{1}{100}$