## Task B-1.1.

For positive real numbers $a$ and $b$, define $a \star b=\frac{a-b}{a+b}$. What is

$$
\frac{(a \star b)+1}{(b \star a)-1}+\frac{(a \star b)+2}{(b \star a)-2}+\ldots+\frac{(a \star b)+k}{(b \star a)-k}+\ldots+\frac{(a \star b)+2020}{(b \star a)-2020}
$$
## Solution.

$$
\begin{gathered}
\frac{(a \star b)+1}{(b \star a)-1}=\frac{\frac{a-b}{a+b}+1}{\frac{b-a}{b+a}-1}=\frac{a-b+(a+b)}{b-a-(b+a)}=\frac{2 a}{-2 a}=-1 \\
\frac{(a \star b)+2}{(b \star a)-2}=\frac{\frac{a-b}{a+b}+2}{\frac{b-a}{b+a}-2}=\frac{a-b+2(a+b)}{b-a-2(b+a)}=\frac{3 a+b}{-b-3 a}=-1 \\
\cdots \\
\frac{(a \star b)+k}{(b \star a)-k}=\frac{\frac{a-b}{a+b}+k}{\frac{b-a}{b+a}-k}=\frac{a-b+k(a+b)}{b-a-k(b+a)}=\frac{(k+1) a+(k-1) b}{(1-k) b-(1+k) a}=-1
\end{gathered}
$$

All addends have a value of -1, which implies that the solution is -2020.

1 point

Note: If the student obtained the correct result without verifying the general case, assign 4 points.