Toán 1/3 – 2/3^2 + 3/3^3 – … – 3/3^100 So sánh với 3/16 10/09/2021 By Elliana 1/3 – 2/3^2 + 3/3^3 – … – 3/3^100 So sánh với 3/16
Giải thích các bước giải: Đặt `A = 1/3 – 2/3^2 + 3/3^3 – … – 3/3^100` `=> 3A = 1 – 2/3 + 3/3^2 + ….. – 3/3^99` `=> 3A + A = (1 – 2/3 + 3/3^2 + ….. – 3/3^99) + (1/3 – 2/3^2 + 3/3^3 – … – 3/3^100)` `=> 4A = 1 – 1/3 + 1/3^2 – 1/3^3 + …. – 1/3^99 + 3/3^100` `=> 4A = B – 3/3^100` Với `B = 1 – 1/3 + 1/3^2 – 1/3^3 + …. – 1/3^99` `=> 3B = 3 – 1 + 1/3 – 1/3^2 + … – 1/3^98` `=> 3B + B = (3 – 1 + 1/3 – 1/3^2 + … – 1/3^98)+( 1 – 1/3 + 1/3^2 – 1/3^3 + …. – 1/3^99)` `=> 4B = 3 – 1/3^99` `=> B = 3/4 – 1/(3^99 . 4) < 3/4` Vậy `4A < 3/4 – 3/3^100 < 3/4` `=> A<3/16 ( Đpcm)` Trả lời
Giải thích các bước giải: Đặt: $A=\dfrac13-\dfrac2{3^2}+\dfrac3{3^3}-…-\dfrac3{3^{100}}$ $\to 3A=1-\dfrac2{3}+\dfrac3{3^2}-…-\dfrac3{3^{99}}$ $\to 3A+A=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+…-\dfrac{1}{3^{99}}-\dfrac3{3^{100}}$ $\to 4A=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+…-\dfrac{1}{3^{99}}-\dfrac3{3^{100}}$ Ta có: $B=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+…-\dfrac{1}{3^{99}}$ $\to 3B=3-1+\dfrac1{3}-\dfrac1{3^2}+…-\dfrac{1}{3^{98}}$ $\to 3B+B=3-\dfrac1{3^{99}}$ $\to 4B=3-\dfrac1{3^{99}}$ $\to B=\dfrac34- \dfrac1{4\cdot 3^{99}}$ $\to 4A=\dfrac34-\dfrac1{4\cdot 3^{99}}-\dfrac3{3^{100}}$ $\to 4A<\dfrac34$ $\to A<\dfrac3{16}$ Trả lời
Giải thích các bước giải:
Đặt `A = 1/3 – 2/3^2 + 3/3^3 – … – 3/3^100`
`=> 3A = 1 – 2/3 + 3/3^2 + ….. – 3/3^99`
`=> 3A + A = (1 – 2/3 + 3/3^2 + ….. – 3/3^99) + (1/3 – 2/3^2 + 3/3^3 – … – 3/3^100)`
`=> 4A = 1 – 1/3 + 1/3^2 – 1/3^3 + …. – 1/3^99 + 3/3^100`
`=> 4A = B – 3/3^100`
Với `B = 1 – 1/3 + 1/3^2 – 1/3^3 + …. – 1/3^99`
`=> 3B = 3 – 1 + 1/3 – 1/3^2 + … – 1/3^98`
`=> 3B + B = (3 – 1 + 1/3 – 1/3^2 + … – 1/3^98)+( 1 – 1/3 + 1/3^2 – 1/3^3 + …. – 1/3^99)`
`=> 4B = 3 – 1/3^99`
`=> B = 3/4 – 1/(3^99 . 4) < 3/4`
Vậy `4A < 3/4 – 3/3^100 < 3/4`
`=> A<3/16 ( Đpcm)`
Giải thích các bước giải:
Đặt:
$A=\dfrac13-\dfrac2{3^2}+\dfrac3{3^3}-…-\dfrac3{3^{100}}$
$\to 3A=1-\dfrac2{3}+\dfrac3{3^2}-…-\dfrac3{3^{99}}$
$\to 3A+A=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+…-\dfrac{1}{3^{99}}-\dfrac3{3^{100}}$
$\to 4A=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+…-\dfrac{1}{3^{99}}-\dfrac3{3^{100}}$
Ta có:
$B=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}+…-\dfrac{1}{3^{99}}$
$\to 3B=3-1+\dfrac1{3}-\dfrac1{3^2}+…-\dfrac{1}{3^{98}}$
$\to 3B+B=3-\dfrac1{3^{99}}$
$\to 4B=3-\dfrac1{3^{99}}$
$\to B=\dfrac34- \dfrac1{4\cdot 3^{99}}$
$\to 4A=\dfrac34-\dfrac1{4\cdot 3^{99}}-\dfrac3{3^{100}}$
$\to 4A<\dfrac34$
$\to A<\dfrac3{16}$