$\frac{1}{3}$-$\frac{1}{3^{2} }$ +$\frac{1}{3^{3} }$ -…-$\frac{1}{3^{100} }$ 10/07/2021 Bởi Alexandra $\frac{1}{3}$-$\frac{1}{3^{2} }$ +$\frac{1}{3^{3} }$ -…-$\frac{1}{3^{100} }$
Đặt `A=1/3-1/(3^2)+1/(3^(3))-….-1/(3^(100)` `3A=1-1/3+1/(3^2)-….-1/(3^(99))` `⇔3A+A=1-1/(3^(100))` `⇔4A=1-1/(3^(100))` `⇔A=1/4-1/(4×3^(100))` Bình luận
Đặt `A= 1/3 – 1/3^2 + 1/3^3 -…- 1/3^100` `1/3 A= 1/3 ( 1/3 – 1/3^2 + 1/3^3-…- 1/3^100)` `1/3 A= 1/3^2 – 1/3^3 + 1/3^4-…-1/3^101` `A+ 1/3 A= 1/3 – 1/3^2 + 1/3^3 -…-1/3^100 + 1/3^2 – 1/3^3 + 1/3^4 -…-1/3^101` `4/3 A = 1/3 – 1/3^101` `A= (1/3 – 1/3^101): 4/3` `A= (1/3 – 1/3^101) . 3/4` `A= 1/4 – 1/(3^100 .4)` Vậy `A= 1/4 – 1/(3^100 .4)` Bình luận
Đặt `A=1/3-1/(3^2)+1/(3^(3))-….-1/(3^(100)`
`3A=1-1/3+1/(3^2)-….-1/(3^(99))`
`⇔3A+A=1-1/(3^(100))`
`⇔4A=1-1/(3^(100))`
`⇔A=1/4-1/(4×3^(100))`
Đặt `A= 1/3 – 1/3^2 + 1/3^3 -…- 1/3^100`
`1/3 A= 1/3 ( 1/3 – 1/3^2 + 1/3^3-…- 1/3^100)`
`1/3 A= 1/3^2 – 1/3^3 + 1/3^4-…-1/3^101`
`A+ 1/3 A= 1/3 – 1/3^2 + 1/3^3 -…-1/3^100 + 1/3^2 – 1/3^3 + 1/3^4 -…-1/3^101`
`4/3 A = 1/3 – 1/3^101`
`A= (1/3 – 1/3^101): 4/3`
`A= (1/3 – 1/3^101) . 3/4`
`A= 1/4 – 1/(3^100 .4)`
Vậy `A= 1/4 – 1/(3^100 .4)`