Question

M=1+(1+2)+(1+2+3)+...+(1+2+3+...+2020)
      1 x 2020+2 x 2019+3 x 2018+...+2020x1

Tính M

Answer 1

Ta có: 1 + ( 1  + 2 ) + ( 1 + 2 + 3 ) + ... + ( 1 + 2 + 3 +...+ 2020) 

= ( 1 + 1 + 1 +... + 1 ) + (2 + 2 +...+ 2 ) + ( 3 + 3+...+ 3 ) + ...+ 2020

Có 2020 số 1 ; 2019 số 2 ; 2018 số 3 ;... ; 1 số 2020 

=  2020 x 1 + 2019 x 2 + 2018 x 3 + ... + 2020x 1

=> \(M=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+...+2020\times1}\)

= \(\frac{1\times2020+2\times2019+...+2020\times1}{1\times2020+2\times2019+...+2020\times1}=1\)