2n:(1+11+2+11+2+3+11+2+3+4+...+11+2+3+...+n)=2020

⇔1+1(1+2).2+1(1+3).3+1(1+4).4+...+1(1+n).n=2n:2020

⇔1+12.3+13.4+14.5+...+1n.(1+n)=n:1010

⇔1+12−13+13−14+14−15+...+1n−1n+1=n:1010

⇔1+12−1n+1=n1010⇔1+12−1n+1=n1010

⇔2(n+1)+(n+1)−12(n+1)=n1010

⇔3n+22(n+1)=n1010

⇔3n+2.1010=2n(n+1)

⇔2n2−n−2020=0