<s>[INST] Find the sum$\pmod{1000}$ 
Find $$1\cdot 2 - 2\cdot 3 + 3\cdot 4 - \cdots + 2015 \cdot 2016 \pmod{1000}$$

I first tried factoring,
$$2(1 - 3 + 6 - 10 + \cdots + 2015 \cdot 1008)$$
I know that $\pmod{1000}$ is the last three digits. 
$$= 2 - 6 + 12 - 20 + 30 - 42 + \cdots + 240$$
that is too complicated. Just hints please!
 [/INST] The given series is: $$\sum_{i=1} ^{1008} (2i-1)(2i)-\sum_{i=1} ^{1007}(2i)(2i+1)$$ which is $$4\times\sum_{i=1} ^{1008} i^2-2\times \sum_{i=1} ^{1008} i-4\times \sum_{i=1} ^{1007} i^2+2\times \sum_{i=1} ^{1007} i$$ which, by cancelling out the common terms becomes, $$4\times 1008^2-2\times 1008=2\times 1008\times (2\times 1008-1)=2\times 1008\times 2015$$ which $\pmod {1000}$ is $$2\times 8\times 15=240.$$
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