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zmudz

  • one year ago

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

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  1. Vocaloid
    • one year ago
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    notice anything special about the numerator and the denominator?

  2. zmudz
    • one year ago
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    it is all odd numbers to 199 and all even numbers to 200

  3. Vocaloid
    • one year ago
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    yes, but more specifically, every number in the denominator is just + 1 greater than every number in the numerator, correct?

  4. zmudz
    • one year ago
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    yes, how does that help though?

  5. zmudz
    • one year ago
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    oh, is the answer 2 then?

  6. zmudz
    • one year ago
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    whoops, that was stupid, i think I know the answer now. thank you!

  7. dumbcow
    • one year ago
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    you could simplify it by using sum formula for arithmetic series \[S = \frac{n (a_1 + a_n)}{2}\]

  8. Vocaloid
    • one year ago
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    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

  9. Vocaloid
    • one year ago
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    ^meant to post that earlier, got sidetracked by another question

  10. zmudz
    • one year ago
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    thanks

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