Cho A=1/2+1/2²+1/2³+….+1/2¹⁰⁰ Chứng minh rằng:0 27/07/2021 Bởi Clara Cho A=1/2+1/2²+1/2³+….+1/2¹⁰⁰ Chứng minh rằng:0 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " Cho A=1/2+1/2²+1/2³+....+1/2¹⁰⁰ Chứng minh rằng:0
$A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+…+\dfrac{1}{2^{100}}$ $2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+…+\dfrac{1}{2^{99}}$ $2A-A=(1+\dfrac{1}{2}+\dfrac{1}{2^2}+…+\dfrac{1}{2^{99}})-(\dfrac{1}{2}+\dfrac{1}{2^2}+…+\dfrac{1}{2^{100}})$ $A=1-\dfrac{1}{2^{100}}$ $A=\dfrac{2^{100}-1}{2^{100}}$ Ta thấy $2^{100}-1<2^{100}$ $→A<1$ mà $2^{100}-1>0,2^{100}>0$ $→\dfrac{2^{100}-1}{2^{100}}>0$ Từ hai điều trên $→0<A<1$ Bình luận
$A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+…+\dfrac{1}{2^{100}}$
$2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+…+\dfrac{1}{2^{99}}$
$2A-A=(1+\dfrac{1}{2}+\dfrac{1}{2^2}+…+\dfrac{1}{2^{99}})-(\dfrac{1}{2}+\dfrac{1}{2^2}+…+\dfrac{1}{2^{100}})$
$A=1-\dfrac{1}{2^{100}}$
$A=\dfrac{2^{100}-1}{2^{100}}$
Ta thấy $2^{100}-1<2^{100}$
$→A<1$
mà $2^{100}-1>0,2^{100}>0$
$→\dfrac{2^{100}-1}{2^{100}}>0$
Từ hai điều trên
$→0<A<1$