# If -f(x) + f(x-1) = 3x and f(1)=1000, what is f(20) ?

Jul 12, 2016

$f \left(20\right) = 373$

#### Explanation:

Given

$- f \left(x\right) + f \left(x - 1\right) = 3 x \ldots \left(1\right) \text{ And } f \left(1\right) = 1000$

$\text{Putting", x=1 " in "(1) "we have}$

$\implies - f \left(1\right) + f \left(1 - 1\right) = 3 \cdot 1$

$\implies - 1000 + f \left(0\right) = 3$

$\implies f \left(0\right) = 1003$

$\text{Now putting "x=1,2,3,...20 " in "(1)" we get}$

$\cancel{- f \left(1\right)} + f \left(0\right) = 3 \cdot 1$

$\cancel{- f \left(2\right)} + \cancel{f \left(1\right)} = 3 \cdot 2$

$\cancel{- f \left(3\right)} + \cancel{f \left(2\right)} = 3 \cdot 3$

$\cancel{- f \left(4\right)} + \cancel{f \left(3\right)} = 3 \cdot 4$
$\ldots \ldots \ldots \ldots \ldots .$
$\ldots \ldots \ldots \ldots \ldots .$
$\ldots \ldots \ldots \ldots \ldots .$

$- f \left(20\right) + \cancel{f \left(19\right)} = 3 \cdot 20$

So adding above last 20 equations we finally get

$- f \left(20\right) + f \left(0\right) = 3 \left(1 + 2 + 3 + \ldots . + 20\right)$

$\implies - f \left(20\right) + 1003 = 3 \cdot \frac{20}{2} \left(20 + 1\right)$

$\implies - f \left(20\right) + 1003 = 630$

$\implies f \left(20\right) = 1003 - 630 = 373$