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

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Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.2notice anything special about the numerator and the denominator?

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0it is all odd numbers to 199 and all even numbers to 200

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.2yes, but more specifically, every number in the denominator is just + 1 greater than every number in the numerator, correct?

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0yes, how does that help though?

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0oh, is the answer 2 then?

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0whoops, that was stupid, i think I know the answer now. thank you!

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0you could simplify it by using sum formula for arithmetic series \[S = \frac{n (a_1 + a_n)}{2}\]

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.2this 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
 one year ago
Best ResponseYou've already chosen the best response.2^meant to post that earlier, got sidetracked by another question
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