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Cho \(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\). Chứng minh A < 2.
\(2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...+\dfrac{100}{2^{99}}\)
=> \(2A-A=A=1+\dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+....+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Đặt \(B=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}\)
=> \(2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{98}}\)
=> \(B=\dfrac{1}{2^2}-\dfrac{1}{2^{99}}\)
=> \(A=1+\dfrac{3}{2^2}+\dfrac{1}{2^2}-\dfrac{100}{2^{100}}-\dfrac{1}{2^{99}}\)
=> \(A=2-\dfrac{102}{2^{100}}< 2\)
Bài 3:
a: a*S=a^2+a^3+...+a^2023
=>(a-1)*S=a^2023-a
=>\(S=\dfrac{a^{2023}-a}{a-1}\)
b: a*B=a^2-a^3+...-a^2023
=>(a+1)B=a-a^2023
=>\(B=\dfrac{a-a^{2023}}{a+1}\)
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)
\(2A-A=1+\frac{2}{2}-\frac{1}{2}+\frac{3}{2^2}-\frac{2}{2^2}+...+\frac{100}{2^{99}}-\frac{99}{2^{99}}-\frac{100}{2^{100}}\)
\(\Rightarrow A=2+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(\Rightarrow A=2.\frac{1}{2^{100}}\)
Vậy \(A< 2\) do \(A=2\) nhân với một phân số nhỏ hơn \(1\)
Ta có:
A=(41+42)+(43+44)+...+(499+4100)
A=4.(1+4)+43.(1+4)+...+499.(1+4)
A=4.5+43.5+...+499.5
A=5.(4+43+...+499)
=>A chia hết cho 5
bài này tớ đã biết nhưng chỉ thử các bạn thôi... cám ơn nhiều nha
a) Mỗi biểu thức M và N đều có 50 thừa số
Ta thấy \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
Vậy \(M< N\)
b) \(M.N=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}\)
\(=\frac{1}{101}\)
c) Vì \(M< N\)nên \(M.M< M.N\)hay \(M.M< \frac{1}{101}< \frac{1}{100}\). Do đó \(M.M< \frac{1}{100}=\frac{1}{10}.\frac{1}{10}\)suy ra \(M< \frac{1}{10}\)( Vì \(M>0\))
\(A=1+4+4^2+......+4^{100}\)
\(A=5+4+4^2+.....+4^{100}\)
\(A=5+4\left(1+4\right)+4^3\left(1+4\right)+......+4^{99}\left(1+4\right)\)
\(A=5+4\cdot5+4^3\cdot5+......+4^{99}\cdot5\)
\(A=5\left(1+4+4^3+.....+4^{99}\right)⋮5\)
Vậy \(A⋮5\)
Ta có : \(VT=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{99}{100!}=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+\frac{5-1}{5!}+...+\frac{100-1}{100!}\)
\(=\frac{2}{1.2}-\frac{1}{2!}+\frac{3}{1.2.3}-\frac{1}{3!}+\frac{4}{1.2.3.4}-\frac{1}{4!}+\frac{5}{1.2.3.4.5}-\frac{1}{5!}+...+\frac{100}{1.2...99.100}-\frac{1}{100!}\)
\(=\frac{1}{1}-\frac{1}{2!}+\frac{1}{1.2}-\frac{1}{3!}+\frac{1}{1.2.3}-\frac{1}{4!}+\frac{1}{1.2.3.4}-\frac{1}{5!}+...+\frac{1}{1.2...99}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{4!}-\frac{1}{5!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
= 4 + 42 + 43 + ... + 4100 chia hết cho 5
= ( 4 + 42 ) + ( 43 + 44 ) + ( 499 + 4100 )
= 4( 1 + 4 ) + 43( 1 + 4 )+ ... + 499( 1 + 4 )
= 4.5+43.5 + ... +499.5
= 5( 4 + 43 + ... + 499 )
\(\Rightarrow\)Dãy trên chia hết cho 5
\(=\left(4+4^2\right)+\left(4^3+4^4\right)+.......+\left(4^{99}+4^{100}\right)\)
\(=4\left(1+4\right)+4^3\left(1+4\right)+..........+4^{99}\left(1+4\right)\)
\(=5\left(4+4^3+..........+4^{99}\right)⋮5\)
\(\Rightarrow4+4^2+4^3+.........+4^{100}⋮5\)