+0

Sum

0
4
2
+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
+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

1000

times, which equals

−1000

. So, the value of the given expression is

−1000

.

Dec 31, 2023
#2
+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

Dec 31, 2023