+0

need help with sum

0
12
1
+1351

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.)

Mar 9, 2024

#1
+394
+3

The function inside, $$f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\right\rfloor$$ is $$\frac{2-3x}{3x+1}$$.

Using what we know about rational functions, set the inside function,  to g(x), $$g(x) = \frac{2-3x}{3x+1}$$ has a x intercept of 2/3, and a horizontal asymptote of -1, and a vertical asymptote at x = -1/3.

Using this information, we can determine the relative shape of the graph.

Because the function has an x intercept at 2/3, and the function only goes down after that, but will not go down past -1, since that is our horizontal asymptote. Plugging in values from $$1\le x \le \infty$$ will always result in a output of $$-1\le g(x) \le 0$$.

Therefore because $$f(x)=\lfloor\frac{2-3x}{3x+1}\rfloor$$, if x is from 1 to 1000 will always result in -1.

Therefore $$f(1)+f(2)+f(3) + \dots + f(999) + f(1000) = -1000$$.

Mar 9, 2024

#1
+394
+3

The function inside, $$f(x) = \left\lfloor\frac{2 - 3x}{3x + 1}\right\rfloor$$ is $$\frac{2-3x}{3x+1}$$.

Using what we know about rational functions, set the inside function,  to g(x), $$g(x) = \frac{2-3x}{3x+1}$$ has a x intercept of 2/3, and a horizontal asymptote of -1, and a vertical asymptote at x = -1/3.

Using this information, we can determine the relative shape of the graph.

Because the function has an x intercept at 2/3, and the function only goes down after that, but will not go down past -1, since that is our horizontal asymptote. Plugging in values from $$1\le x \le \infty$$ will always result in a output of $$-1\le g(x) \le 0$$.

Therefore because $$f(x)=\lfloor\frac{2-3x}{3x+1}\rfloor$$, if x is from 1 to 1000 will always result in -1.

Therefore $$f(1)+f(2)+f(3) + \dots + f(999) + f(1000) = -1000$$.

hairyberry Mar 9, 2024