Consecutive is (not so) unpredictable (Part 3)

The given expression can be written as ,

$1-(2+3+4+5+...................+999+1000.)$

Adding and subtracting $1.$

$1-[(1+2+3+4+5+..............+999+1000)-1.]$

$1-[\large\sum_{n=1}^{1000} -1].$

$1-[\dfrac{1000*1001}{2}-1].$

$1-[500500-1.]$

$1-[500499].$

$=-500498.$

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$A=1-2-3-\ldots-999-1000$

$A=(2-1)-2-3-\ldots-999-1000$

$A= 2-(1+2+3+\ldots+999+1000)$

$2A= 4-(1001+1001+1001+\ldots+1001+1001)$

$2A = 4 - 1001 \times 1000$

$\Rightarrow A=2-\frac{1001\times1000}{2}$

$A= 2-500500 = \boxed{-500498}$