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- zmudz

Simplify (1+3+5+...+199)/(2+4+6+...+200). Thanks!

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- zmudz

Simplify (1+3+5+...+199)/(2+4+6+...+200). Thanks!

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- Vocaloid

notice anything special about the numerator and the denominator?

- zmudz

it is all odd numbers to 199 and all even numbers to 200

- Vocaloid

yes, but more specifically, every number in the denominator is just + 1 greater than every number in the numerator, correct?

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- zmudz

yes, how does that help though?

- zmudz

oh, is the answer 2 then?

- zmudz

whoops, that was stupid, i think I know the answer now. thank you!

- dumbcow

you could simplify it by using sum formula for arithmetic series
\[S = \frac{n (a_1 + a_n)}{2}\]

- Vocaloid

this probably won't be read by anyone, but eh...
we have 100 terms in the numerator and 100 terms in the denominator
we can calculate the sum 1 + 3 + ... 199 using the formula above, and represent this sum as S
if we let the numerator equal S, then the denominator is S + 100, since there are 100 terms in the denominator and each term is 1 greater than its corresponding term in the numerator
so our answer is something along the lines of S/(S+100) which allows use to use the sum formula once instead of twice

- Vocaloid

^meant to post that earlier, got sidetracked by another question

- zmudz

thanks

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