`Đặt A = 1/3 + 1/6 + 1/10 + ....... + 1/(1 + 2 + ........... + 2015)`
`A = 2/6 + 2/12 + 2/20 + .......... + 2/(2(1 + 2 + ........... + 2015))`
`A = 2(1/6 + 1/12 + 1/20 + ............. + 1/(2(1 + 2 + ........... + 2015)))`
`Đặt B = 2(1 + 2 + .......... + 2015)`
`text(B có số số hạng là :)`
`(2015 - 1) : 1 + 1 = 2015 (số)`
`B = 2 . 2015 . (2015 + 1) : 2`
`B = 2015 . 2016`
`A = 2(1/(2 . 3) + 1/(3 . 4) + 1/(4 . 5) + ............. + 1/(2015 . 2016))`
`A = 2(1/2 - 1/3 + 1/3 - 1/4 + ................. + 1/2015 - 1/2016)`
`A = 2(1/2 - 1/2016)`
`A = 1 - 1/1008`
`A = 1007/1008`