Question

I'm bored. Time for some math problems.

Answer 1

Gimme a sec.

Answer 2

Lemme search for a good one.

Answer 3

Feel free to tag people. Just not too many, pwease.

Answer 4

these look easy but ever since i took a knee to my arrow....

Answer 5

LOOL

Answer 6

solution for #5 incoming

Answer 7

the bottom can be factored as (n-1)(n+1). from there you can cancels things accordingly.
IE
\[2^2/(2^2 -1)* 3^2/(3^2-1) = 2^2/[(2-1)(2+1)] * 3^2/[(3-1)(3+1)]\]
also notice that you can just cancel one of the numbers before and after, IE (2+1) will cancel one of the 3's, and (3-1) will cancel one of the 2's.
This is the genesis of this question.

you will end up having:
\[99!/100\]

Answer 8

You're on the right track, but teh answer is incorrect.

Answer 9

oh. The only number left on the numerator are 2 * 99, since these numbers only get canceled once since they are on the edges. so the actual soultion is

\[2*99/100 = 99/50\]

i hope D:

Answer 10

Yup.

Answer 11

@FoolForMath , wanna join? THese are easy, but still.

Answer 12

lol. Maybe they can explain telescoping series to me and how to solve them quicker.

Answer 13

lool. These are 7th grade math probs.

Answer 14

For Q5, is it2(99)/100 => 99/50?

Answer 15

99/50.

Answer 16

so, yes.

Answer 17

Cool~~~ Should I show workings?

Answer 18

Well, i'm pretty sure there are only 3 ways to solve this.

Answer 19

1. What alexander did (prob what you did)
2. number crunching LOOOL
3. Capital pi and some fancy shtuff

Answer 20

For Q4, is it 11/20?

Answer 21

the solution for #4 is pretty much the same, however the binomial is on the numerator this time. The same strategy should work.

Answer 22

For Q6, is it 2?

Answer 23

I wish I had my monster geometry problems on me. These are too eas.