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Let
\[f(x) = \left\lfloor\frac{2 - 3x}{2x + 8}\right\rfloor.\]
Evaluate $f(1)+f(2) + f(3) + \dots + f(999)+f(1000).$ (This sum has $1000$ terms, one for the result when we input each integer from $1$ to $1000$ into ${}f$.)

 Sep 7, 2023
 #1
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Note

 

f(1)  = floor [  -1/10 ]  =   -1   

f(2)  =  floor[  -4/12 ] =   -1

f(3)  =  floor [  -7 / 14 ]   = -1

f(4)  =  floor [  -10 / 16 ] = -1

.

.

f(10) = floor [-28 /28] = -1 

f(11) = floor [  -31 / 30]  = -2

.

.

f(1000)  = floor [ -2998 / 2008 ] = - 2

 

We have that the first 10 terms evaluate to  -1    and the  last 990 evaluate to -2

 

Sum  = 10 (-1) + 990 ( -2)  = -1990

 

 

cool cool cool

 Sep 8, 2023

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