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Let f(x) = \left\lfloor\dfrac{2 - 3x}{x + 5}\right\rfloor. Evaluatef(1)+f(2) + f(3) + \cdots + f(999)+f(1000).(This sum has 1000 terms, one for the result when we input each integer from 1 to 1000 into f.)

Guest Jan 8, 2015

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 #1
avatar+78575 
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floor [ (2-3x) / (x + 5) ]

This isn't as daunting as it seems

Notice the first few terms....

f(1) = floor( -1 /6) = -1   ..   f(2) = floor (-4/7) = -1... f(3) = floor (-7/8) = -1  ...f(4) = floor (-10/9)= -2

And f(11)  = floor (-31/16) = -2   ....and f(12) = floor (-34 / 17) = -2 and f(13) =floor (-37/18) = -3

And every successive term has has a floor of -3....for example f(1000) = floor(-2998/ 1005) = -3

So.... 3 terms have a floor = -1

And 9 terms have a floor of -2

And (1000 - 12) = (988) terms have a floor of -3

So the sum is   3(-1) + 9(-2) + 988(-3)  = -2985

 

CPhill  Jan 8, 2015
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 #1
avatar+78575 
+10
Best Answer

floor [ (2-3x) / (x + 5) ]

This isn't as daunting as it seems

Notice the first few terms....

f(1) = floor( -1 /6) = -1   ..   f(2) = floor (-4/7) = -1... f(3) = floor (-7/8) = -1  ...f(4) = floor (-10/9)= -2

And f(11)  = floor (-31/16) = -2   ....and f(12) = floor (-34 / 17) = -2 and f(13) =floor (-37/18) = -3

And every successive term has has a floor of -3....for example f(1000) = floor(-2998/ 1005) = -3

So.... 3 terms have a floor = -1

And 9 terms have a floor of -2

And (1000 - 12) = (988) terms have a floor of -3

So the sum is   3(-1) + 9(-2) + 988(-3)  = -2985

 

CPhill  Jan 8, 2015

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