P= $\frac{1}{100}$ – $\frac{1}{100×99}$ – $\frac{1}{99×98}$ – … – $\frac{1}{3×2}$ – $\frac{1}{2×1}$ 03/07/2021 Bởi Lyla P= $\frac{1}{100}$ – $\frac{1}{100×99}$ – $\frac{1}{99×98}$ – … – $\frac{1}{3×2}$ – $\frac{1}{2×1}$
`P = 1/100 – 1/(100.99) – 1/(99.98) – …..- 1/(3.2) – 1/(2.1)` `= 1/100 – (1/(100.99) + 1/(99.98)+ …..- 1/(3.2)+ 1/(2.1))` `= 1/100 – (1/(1.2) + 1/(2.3) + ….+ 1/(98.99) + 1/(99.100))` `= 1/100 – (1 – 1/2 + 1/2 – 1/3+….+1/98-1/99+1/99-1/100)` `= 1/100 – (1 – 1/100) ` `= 1/100 – 99/100` `= -98/100` `= -49/50` (Chúc bạn học tốt) Bình luận
`P = 1/100 – 1/(100.99) – 1/(99.98) – … – 1/(3.2) – 1/(2.1)` `P = 1/100 – (1/(100.99) + 1/(99.98) + …. + 1/(3.2) + 1/(2.1))` Đặt: `A = 1/(100.99) + 1/(99.98) + …. + 1/(3.2) + 1/(2.1)` `A = 1/(1.2) + 1/(2.3) + …. + 1/(98.99) + 1/(99.100)` `A = 1 – 1/2 + 1/2 – 1/3 + 1/3 – …. + 1/99 – 1/100` `A = 1 – 1/100` `A = 99/100` `=> P = 1/100 – (1/(100.99) + 1/(99.98) + …. + 1/(3.2) + 1/(2.1))` `P = 1/100 – 99/100` `P = (-49)/50` Bình luận
`P = 1/100 – 1/(100.99) – 1/(99.98) – …..- 1/(3.2) – 1/(2.1)`
`= 1/100 – (1/(100.99) + 1/(99.98)+ …..- 1/(3.2)+ 1/(2.1))`
`= 1/100 – (1/(1.2) + 1/(2.3) + ….+ 1/(98.99) + 1/(99.100))`
`= 1/100 – (1 – 1/2 + 1/2 – 1/3+….+1/98-1/99+1/99-1/100)`
`= 1/100 – (1 – 1/100) `
`= 1/100 – 99/100`
`= -98/100`
`= -49/50`
(Chúc bạn học tốt)
`P = 1/100 – 1/(100.99) – 1/(99.98) – … – 1/(3.2) – 1/(2.1)`
`P = 1/100 – (1/(100.99) + 1/(99.98) + …. + 1/(3.2) + 1/(2.1))`
Đặt:
`A = 1/(100.99) + 1/(99.98) + …. + 1/(3.2) + 1/(2.1)`
`A = 1/(1.2) + 1/(2.3) + …. + 1/(98.99) + 1/(99.100)`
`A = 1 – 1/2 + 1/2 – 1/3 + 1/3 – …. + 1/99 – 1/100`
`A = 1 – 1/100`
`A = 99/100`
`=> P = 1/100 – (1/(100.99) + 1/(99.98) + …. + 1/(3.2) + 1/(2.1))`
`P = 1/100 – 99/100`
`P = (-49)/50`