7.6. Consider the numbers: $a=2^{2}+4^{2}+6^{2}+\cdots+2020^{2} ; b=1^{2}+3^{2}+5^{2}+\cdots+2019^{2}$; $c=1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+2019 \cdot 2020$. Calculate the value of the expression $E=\frac{a-b}{a+b-c}$.
## Solution:

$a-b=2^{2}+4^{2}+6^{2}+\cdots+2020^{2}-\left(1^{2}+3^{2}+5^{2}+\cdots+2019^{2}\right)=\left(2^{2}-1^{2}\right)+\left(4^{2}-3^{2}\right)+$ $\left(6^{2}-5^{2}\right)+\ldots .+\left(2020^{2}-2019^{2}\right)=3+7+11+\ldots .+4039=4042 \cdot 505$. $a+b-c=1^{2}+2^{2}+3^{2}+$. . $+2020^{2}-(1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots .+2019 \cdot 2020)=1+$ $\left(2^{2}-1 \cdot 2\right)+\left(3^{2}-2 \cdot 3\right)+\left(4^{2}-3 \cdot 4\right)+\ldots+\left(2020^{2}-2019 \cdot 2020\right)=1+2+$ $3+$. . . $+2020=2021 \cdot 1010=4042 \cdot 505$.

Therefore, $E=1$.

Answer: $E=1$.