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Let \(f(x) = \left\lfloor\dfrac{2 - 3x}{x + 5}\right\rfloor\). Evaluate \(f(1)+f(2) + f(3) + \cdots + f(999)+f(1000).\)

 Apr 7, 2020
 #1
avatar+932 
0

Edit: Wrong Answer

 

 

f(1)= -1 and f(1000)= -3. The sum of the terms is equal to the average multiplied by the number of terms, so 1000(-1+-3)/2= -2000.

 

*Check me if I'm wrong

 

Hope it helps!

 Apr 7, 2020
edited by HELPMEEEEEEEEEEEEE  Apr 7, 2020
 #2
avatar+111329 
+2

Look at  the  graph here...https://www.desmos.com/calculator/nofaug4qzu

 

And note  that  

 

f(1)  =  -1       f(4)  = -2     f(7)  = -2     f(10)  = -2

f(2)  = -1        f(5) = -2      f(8) = -2      f(11)  =  -2

f(3)  = -1        f(6) = -2      f(9)  = -2     f(12)  = -2

 

f(13)  = -3

...

f(1000) = -3

 

This  implies that  the sum is   (-1)*3 + (-2)*9  + (1000 - 13 + 1) (-3)   =  -2985

 

 

cool cool cool

 Apr 7, 2020

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