Bài làm :
Ta có :
\(x+2x+3x+...+2020x=2020.2021\)
\(\Leftrightarrow x\left(1+2+3+...+2020\right)=2020.2021\)
\(\Leftrightarrow x.\frac{\left(2020+1\right).2020}{2}=2021.2020\)
\(\Leftrightarrow x.\frac{2021.2020}{2}=2021.2020\)
\(\Leftrightarrow x=2\)
Vậy x=2
\(x+2x+3x+...+2020x=2020\cdot2021\)
\(x\left(1+2+3+...+2020\right)=2020\cdot2021\)
1 + 2 + 3 ... + 2020
Số số hạng :
\(\left(2020-1\right):1+1=2020\)
Tổng :
\(\left(2020+1\right)\cdot2020:2=2021\cdot1010\)
\(2021\cdot1010\cdot x=2020\cdot2021\)
\(1010x=2020\)
\(x=2\)
