6. For all non-negative values of the real variable $x$, the function $f(x)$ satisfies the condition $f(x+1)+1=f(x)+\frac{43}{(x+1)(x+2)}$. Calculate $\frac{101}{f(2020)}$, if $\quad f(0)=2020$.
# Solution.

Notice that

$f(x+2020)-f(x)=(f(x+2020)-f(x+2019))+(f(x+2019)-f(x+2018))+\cdots+(f(x+$ 1) $-f(x))=\frac{43}{(x+2020)(x+2021)}-1+\frac{43}{(x+2019)(x+2020)}-1+\cdots+\frac{43}{(x+1)(x+2)}-1$. Thus, $f(2020)-f(0)=43\left(\frac{1}{2020}-\frac{1}{2021}+\cdots+1-\frac{1}{2}\right)-2020=43\left(1-\frac{1}{2021}\right)-2020 . \quad$ Therefore, $\frac{101}{f(2020)}=\frac{47}{20}=2.35$.

Answer: 2.35 .