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Let f(x) = \left \lfloor \frac{2 - 3x}{x + 3} + x^2 \right \rfloor.
Find f(1) + f(2) + f(3) + ... + f(999) + f(1000).

 May 17, 2024
 #1
avatar+129514 
+1

\( \left \lfloor \frac{2 - 3x}{x + 3} + x^2 \right \rfloor \)

 

f(1)  = 0

f(2)  = 3

f(3) = 7

f(4)  = 14

f(5)  = 23

f(6)  = 34

f(7) = 47

f(8)  = 62

f(9) = 78

.

.

.

f (1000)  =  999,997

 

It appears that, from x = 3 to x = 8 inc;usive,  f(x)  =  x^2 - 2

And from x = 9  to x =1000 inclusive , f(x)  = x^2 - 3

 

So

 

 

 

So....the sum is

 

         8                      1000

3  +   Σ   x^2  - 2   +   Σ  x^2 - 3      =  333,830,510

       x = 3                 x = 9

 

cool cool cool

 May 17, 2024

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