Đặt :
\(H=1^2-2^2+3^2-4^2+5^2-6^2+......+2019^2-2020^2\)
\(=\left(1^2-2^2\right)+\left(3^2-4^2\right)+.\left(5^2-6^2\right)+...+\left(2019^2-2020^2\right)\) (Có 1010 nhóm)
\(=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+....+\left(2019-2020\right)\left(2019+2020\right)\)
\(=-3-7-11-......-4039\)
\(=-\left(3+7+11+4039\right)\)
\(=-\frac{\left(4039+3\right).1010}{2}\)
\(=-2041210\)
Vậy....
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