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avatar+794 

Let

f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\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.)

 Dec 31, 2023
 #1
avatar+6 
+1

The function you’ve given is

f(x)=⌊3x+12−3x​⌋

where the notation

⌊y⌋

represents the greatest integer less than or equal to

y

.

Let’s evaluate this function for a few values:

For

x=1

,

f(1)=⌊3(1)+12−3(1)​⌋=⌊4−1​⌋=−1

because the greatest integer less than

−1/4

is

−1

.For

x=2

,

f(2)=⌊3(2)+12−3(2)​⌋=⌊7−4​⌋=−1

because the greatest integer less than

−4/7

is

−1

.

You can see a pattern here. The function

f(x)

is always equal to

−1

for all positive integers

x

. This is because the numerator

2−3x

is always negative and the denominator

3x+1

is always positive for all positive integers

x

. The fraction is thus a negative number between

0

and

−1

, and the floor of this number is always

−1

.

Therefore, the sum

f(1)+f(2)+f(3)+⋯+f(999)+f(1000)

is simply

−1

added to itself

1000

times, which equals

−1000

. So, the value of the given expression is

−1000

.

 Dec 31, 2023
 #2
avatar+126665 
+1

\(f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\right\rfloor \)

 

Write this  in another way

 

floor  [  (-3x + 2) / (3x + 1) ] 

 

Evaluating this  function (without the floor) at f(1)  = -1/4    and floor (-1/4)  = -1

 

And any positive integer  value of  x produces a function value (without the floor)   which is greater than -1  but less than 0  

 

So the floor of all such values =  -1

 

So

 

f(1)  + f(2)  +......+ f(999) + f(1000)   =

 

1000 (-1)  =   -1000

 

cool cool cool

 Dec 31, 2023

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