B=2×2012/(1+1/(1+2)+1/(1+2+3)+…+1/(1+2+3+4+…+2012)) tính giá trị B

`B=2.2012/{1+1/{1+2}+1/{1+2+3}+…+1/{1+2+3+4+…+2012}}`

Đặt `C=1+1/{1+2}+1/{1+2+3}+…+1/{1+2+3+4+…+2012}`

Ta có công thức: `1+2+3+…+n={n.(n+1)}/2`

Thay vào `C`
`⇒C=1+1/{{2.3}/2}+1/{{3.4}/2}+…+1/{{2012.2013}/2}`

`⇒C/2=1/2+1/2.3+1/3.4+…+1/2012.2013`

`⇒C/2=1/2+1/2-1/3+1/3-1/4+…+1/2012-1/2013`

`⇒C/2=1-1/2013`

`⇒C/2=2012/2013`

`⇒C=4024/2013`

`⇒B=4024/{4024/2013}`

`⇒B=2013`