Toán Chứng minh B < 1 1/2 + (1/2)^2 + (1/2)^3+....+ (1/2)^99 23/07/2021 By Jade Chứng minh B < 1 1/2 + (1/2)^2 + (1/2)^3+....+ (1/2)^99
Đáp án + Giải thích các bước giải: `B=1/2 + (1/2)^2 + (1/2)^3+…+ (1/2)^99``=>B=1/2+1/2^2+1/2^3+…+1/2^99``=>2B=1+1/2+1/2^2+…+1/2^98``=>2B-B=(1+1/2+1/2^2+…+1/2^98)-(1/2+1/2^2+1/2^3+…+1/2^99)``=>B=1-1/2^99<1`Vậy `B<1` Trả lời
Ta có: $B = \dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{99}}$ $\Leftrightarrow 2B = 1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{98}}$ $\Leftrightarrow 2B – B = \left(1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{98}}\right) – \left(\dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{99}}\right)$ $\Leftrightarrow B = 1 – \dfrac{1}{2^{99}} < 1$ Trả lời
Đáp án + Giải thích các bước giải:
`B=1/2 + (1/2)^2 + (1/2)^3+…+ (1/2)^99`
`=>B=1/2+1/2^2+1/2^3+…+1/2^99`
`=>2B=1+1/2+1/2^2+…+1/2^98`
`=>2B-B=(1+1/2+1/2^2+…+1/2^98)-(1/2+1/2^2+1/2^3+…+1/2^99)`
`=>B=1-1/2^99<1`
Vậy `B<1`
Ta có: $B = \dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{99}}$
$\Leftrightarrow 2B = 1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{98}}$
$\Leftrightarrow 2B – B = \left(1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \cdots + \dfrac{1}{2^{98}}\right) – \left(\dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^{99}}\right)$
$\Leftrightarrow B = 1 – \dfrac{1}{2^{99}} < 1$