<s>[INST] How to describe 4-8+12-16+20-24+28 using summation ($\Sigma$) notation? Can anyone find the ∑ summation for this please?
4-8+12-16+20-24+28
It seems to be going up by steps of 4, but I can't seem to get how I should write it down, since it uses both + and -
 [/INST] $$4-8+12-16+20-24+28=4(1-2+3-4+5-6+7)=4((-1)^{1+1}\cdot 1+(-1)^{2+1}\cdot 2+(-1)^{3+1}\cdot 3+(-1)^{4+1} \cdot 4+(-1)^{5+1} \cdot 5+(-1)^{6+1} \cdot 6+(-1)^{7+1} \cdot 7)=4 \sum_{i=1}^{7}(-1)^{i+1}i=4 \left ( \sum_{i=0}^{3} (2i+1)-\sum_{i=1}^{3} 2i\right )=4 \left( 2 \sum_{i=0}^{3} i +\sum_{i=0}^{3} 1-2\sum_{i=1}^{3} i\right )=4 \left( 2 \sum_{i=1}^{3} i +\sum_{i=0}^{3} 1-2\sum_{i=1}^{3} i\right ) \\ =4 \sum_{i=0}^{3} 1=4 \cdot (3+1)=4 \cdot 4=16$$
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