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Cho A= 1/2 + 2/2^2 + 3/2^3 + .... + 99/2^99 + 100/2^100
So sánh A với 2
Xl mấy bạn mk gõ thiếu bài gốc đây nhá Giải giúp mk với nạ
\(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)
\(2A=1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\)
\(2A-A=\left(1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\right)\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Đặt:
\(B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(2B=2+1+\dfrac{1}{2^2}+....+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\)
\(2B-B=\left(2+1+\dfrac{1}{2^2}+....+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(B=2-\dfrac{1}{2^{99}}\)
Vậy \(A=2-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}< 2\)
Bây giờ mình đang bận học bài 1 chút.Xíu nữa mình làm cho nhé
\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{99.100}\)
Đặt B \(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(\Rightarrow B< 1\)
\(\Rightarrow S< 1\)
S=\(\left(2+2^2\right)+\left(2^3+2^4\right)\)+......+\(\left(2^{99}+2^{100}\right)\)
=2(
a, S = 2 + 22 + 23 + 24 + ... + 299 + 2100. 2S = 22 + 23 + 24 + 25 + ... + 2100 + 2101 => 2S - S = S = (22 + 23 + 24 + 25 + ... + 2100 + 2101) - (2 + 22 + 23 + 24 + ... + 299 + 2100) = 2101 - 2. Vậy S = 2101 - 2. b, S = 2 + 22 + 23 + 24 + ... + 299 + 2100 = (2 + 22) + (23 + 24) + ... + (299 + 2100) = 2.(1 + 2) + 23.(1 + 2) + ... + 299.(1 + 2) = (1 + 2).(2 + 23 + ... + 299) = 3.(2 + 23 + ... + 299) => S ⋮ 3. Vậy S ⋮ 3 (đpcm)
a) \(S=2+2^2+2^3+...+2^{100}\)
\(S=\left(2+2^2+2^3+2^4\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(S=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
\(S=2\cdot15+...+2^{97}\cdot15\)
\(S=15\cdot\left(2+...+2^{97}\right)⋮15\left(đpcm\right)\)
b) \(S=2+2^2+...+2^{100}\)
\(2S=2^2+2^3+...+2^{101}\)
\(2S-S=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(S=2^{101}-2< 2^{101}\)
s=2+2^2+2^3+......÷2^100=> 2s=2^2+2^3+2^4+.....+2^101
=>2s-s=s=2^101-2=>s<2^101