Toán 1/1×5 + 1/5×10 + 1/10×15 +…+1/2010×2015 01/07/2021 By Autumn 1/1×5 + 1/5×10 + 1/10×15 +…+1/2010×2015
Đặt: `A = 1/(1.5) + 1/(5.10) + 1/(10.15) + … + 1/(2010.2015)` `A = 1/(1.5) + 1/5 . 5 . (1/(5.10) + 1/(10.15) + … + 1/(2010.2015))` `A = 1/(1.5) + 1/5 . (5/(5.10) + 5/(10.15) + … + 5/(2010. 2015))` `A = 1/5 + 1/5 . (1/5 – 1/10 + 1/10 – 1/15 + … + 1/2010 – 1/2015)` `A = 1/5 + 1/5 . (1/5 – 1/2015)` `A = 1/5 . (1 + 1/5 – 1/2015)` `A = 1/5 . 2417/2015` `A = 2417/ 10075` Trả lời
Đáp án: `1/(1xx5)+\underbrace{1/(5xx10)+1/(10xx15)+….+1/(2010xx2015)}_{B}` Ta có: `5B=5/(5xx10)+5/(10xx15)+….+5/(2010xx2015)` `=>5B=1/5-1/10+1/10-1/15+….+1/2010-1/2015` `=>5B=1/5-1/2015` `=>5B=402/2015` `=>B=402/10075` `=>1/5+B=2417/10075` Trả lời
Đặt:
`A = 1/(1.5) + 1/(5.10) + 1/(10.15) + … + 1/(2010.2015)`
`A = 1/(1.5) + 1/5 . 5 . (1/(5.10) + 1/(10.15) + … + 1/(2010.2015))`
`A = 1/(1.5) + 1/5 . (5/(5.10) + 5/(10.15) + … + 5/(2010. 2015))`
`A = 1/5 + 1/5 . (1/5 – 1/10 + 1/10 – 1/15 + … + 1/2010 – 1/2015)`
`A = 1/5 + 1/5 . (1/5 – 1/2015)`
`A = 1/5 . (1 + 1/5 – 1/2015)`
`A = 1/5 . 2417/2015`
`A = 2417/ 10075`
Đáp án:
`1/(1xx5)+\underbrace{1/(5xx10)+1/(10xx15)+….+1/(2010xx2015)}_{B}`
Ta có:
`5B=5/(5xx10)+5/(10xx15)+….+5/(2010xx2015)`
`=>5B=1/5-1/10+1/10-1/15+….+1/2010-1/2015`
`=>5B=1/5-1/2015`
`=>5B=402/2015`
`=>B=402/10075`
`=>1/5+B=2417/10075`