tính : 1+1/2+(1++2)+1/3(1+2+3)+1/4(1+2+3+4)+…+1/20(1+2+3+…+20) 16/11/2021 Bởi Peyton tính : 1+1/2+(1++2)+1/3(1+2+3)+1/4(1+2+3+4)+…+1/20(1+2+3+…+20)
`A= 1+1/2 × (1+2)+1/3 ×(1+2+3)+…+1/20 × (1+2+3+…+20)` `<=> A = 1+ 1/2 × 3 +…+1/20 × 210` `<=> A =1+1,5+2+…+10,5` `<=> A = { (10,5 + 1).[(10,5 – 1) : 0,5 + 1] }/2` `⇔ A= 115` $@Race2k53$ Xin hay nhất ! Bình luận
Ta có: ` A= 1+1/2 . (1+2)+1/3 . (1+2+3)+…+1/20 . (1+2+3+…+20) ` ` <=> A = 1+ 1/2 . 3 +…+1/20 . 210 ` ` <=> A =1+1,5+2+…+10,5 ` ` <=> A = \frac{ (10,5 + 1).[(10,5 – 1) : 0,5 + 1] }{2} ` ` <=> A = \frac{230}{2} ` ` <=> A = 115 ` Bình luận
`A= 1+1/2 × (1+2)+1/3 ×(1+2+3)+…+1/20 × (1+2+3+…+20)`
`<=> A = 1+ 1/2 × 3 +…+1/20 × 210`
`<=> A =1+1,5+2+…+10,5`
`<=> A = { (10,5 + 1).[(10,5 – 1) : 0,5 + 1] }/2`
`⇔ A= 115`
$@Race2k53$
Xin hay nhất !
Ta có:
` A= 1+1/2 . (1+2)+1/3 . (1+2+3)+…+1/20 . (1+2+3+…+20) `
` <=> A = 1+ 1/2 . 3 +…+1/20 . 210 `
` <=> A =1+1,5+2+…+10,5 `
` <=> A = \frac{ (10,5 + 1).[(10,5 – 1) : 0,5 + 1] }{2} `
` <=> A = \frac{230}{2} `
` <=> A = 115 `