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\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{2^{99}}\)
\(=>B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)\)
\(B=1-\dfrac{1}{2^{99}}< 1=>B< 1\)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)
\(B=1-\dfrac{1}{2^{100}}< 1\left(đpcm\right)\)
Vậy \(B< 1\)
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
Vậy...
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
Đặt $A=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{100}}$
Ý bạn là muốn CMR $A<B$?
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Lời giải:
$2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}$
Trừ theo vế:
$2A-A=1-\frac{1}{2^{100}}< 1<2$
$\Leftrightarrow A< 2$ hay $A< B$