Question 20
Find the value of
$$\frac{1-(1/2019)^2}{1+(1/2019)^2}+\frac{1-(1/2018)^2}{1+(1/2018)^2}+\cdots+\frac{1-(1/2)^2}{1+(1/2)^2}$$
$$+\frac{1-1^2}{1+1^2}+\frac{1-2^2}{1+2^2}+\cdots+\frac{1-2019^2}{1+2019^2}.$$
Answer format: an integer without units. Leading zeros are optional; the source uses four digit boxes.
$$\frac{1-(1/2019)^2}{1+(1/2019)^2}+\frac{1-(1/2018)^2}{1+(1/2018)^2}+\cdots+\frac{1-(1/2)^2}{1+(1/2)^2}$$
$$+\frac{1-1^2}{1+1^2}+\frac{1-2^2}{1+2^2}+\cdots+\frac{1-2019^2}{1+2019^2}.$$
Answer format: an integer without units. Leading zeros are optional; the source uses four digit boxes.