Toán $\frac{1}{3}$ +$\frac{1}{3^{2} }$+$\frac{1}{3^{3} }$+…+ $\frac{1}{3^{100}}$ 01/07/2021 By Lyla $\frac{1}{3}$ +$\frac{1}{3^{2} }$+$\frac{1}{3^{3} }$+…+ $\frac{1}{3^{100}}$
Đáp án: `1/2-1/(2.3^100)` Giải thích các bước giải: `A=1/3+1/3^2+1/3^3+…+1/3^100` `=>3A=3.(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+1/3^2+…+1/3^99)-(1/3+1/3^2+1/3^3+…+1/3^100)` `=>2A=1-1/3^100` `=>`$A=\dfrac{1-\dfrac{1}{3^{100}}}{2}$ `=>A=1/2-1/(2.3^100)` Vậy biểu thức có giá trị là `1/2-1/(2.3^100)`. Trả lời
Đặt `A = 1/3 + 1/3^2 + 1/3^3 + …. + 1/3^100` `A . 3 = 1 + 1/3 + 1/3^2 + …. + 1/3^99` `A . 3 – A = ( 1 + 1/3 + 1/3^2 + …. + 1/3^99 ) – ( 1/3 + 1/3^2 + 1/3^3 + …. + 1/3^100 )` `A . 2 = 1 – 1/3^100` `A . 2 = ( 3^100 – 1 )/3^100` `A = ( 3^100 – 1 )/3^100 : 2` `A = ( 3^100 – 1 )/( 3^100 . 2 )` Trả lời
Đáp án:
`1/2-1/(2.3^100)`
Giải thích các bước giải:
`A=1/3+1/3^2+1/3^3+…+1/3^100`
`=>3A=3.(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+1/3^2+…+1/3^99)-(1/3+1/3^2+1/3^3+…+1/3^100)`
`=>2A=1-1/3^100`
`=>`$A=\dfrac{1-\dfrac{1}{3^{100}}}{2}$
`=>A=1/2-1/(2.3^100)`
Vậy biểu thức có giá trị là `1/2-1/(2.3^100)`.
Đặt `A = 1/3 + 1/3^2 + 1/3^3 + …. + 1/3^100`
`A . 3 = 1 + 1/3 + 1/3^2 + …. + 1/3^99`
`A . 3 – A = ( 1 + 1/3 + 1/3^2 + …. + 1/3^99 ) – ( 1/3 + 1/3^2 + 1/3^3 + …. + 1/3^100 )`
`A . 2 = 1 – 1/3^100`
`A . 2 = ( 3^100 – 1 )/3^100`
`A = ( 3^100 – 1 )/3^100 : 2`
`A = ( 3^100 – 1 )/( 3^100 . 2 )`