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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 6: Chứng minh rằng A=1/2 – 2/2^2 + 3/2^3 – 4/2^4 + …99/2^99 – 100/2^100 < 2/9

Toán Lớp 6: Chứng minh rằng
A=1/2 – 2/2^2 + 3/2^3 – 4/2^4 + …99/2^99 – 100/2^100 < 2/9

Comments ( 2 )

  1. \(A=\frac{1}{2}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+…+\frac{99}{2^{99}}-\frac{100}{2^{100}}\)
    \(\Rightarrow2A=1-\frac{2}{2}+\frac{3}{2^2}-\frac{4}{2^3}+\frac{5}{2^4}-\frac{6}{2^5}+\frac{7}{2^6}-…+\frac{99}{2^{98}}-\frac{100}{2^{99}}\)
    Cộng vế theo vế ta được:
    \(3A=1+\left(\frac{1}{2}-\frac{2}{2}\right)+\left(-\frac{2}{2^2}+\frac{3}{2^2}\right)+\left(\frac{3}{2^3}-\frac{4}{2^3}\right)+\left(-\frac{4}{2^4}+\frac{5}{2^4}\right)+…+\left(\frac{99}{2^{99}}-\frac{100}{2^{99}}\right)-\frac{100}{2^{100}}\)
    \(\Rightarrow3A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}+…+\frac{1}{2^{98}}-\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
    Xét \(A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
    \(\Rightarrow2B=2-1+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+…+\frac{1}{2^{97}}-\frac{1}{2^{98}}\)
    Cộng vế theo vế ta được:
    \(3A=2+\left(1-1\right)+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+…+\left(\frac{1}{2^{98}}-\frac{1}{2^{98}}\right)-\frac{1}{2^{99}}\)
    \(\Rightarrow3B=2-\frac{1}{2^{99}}< 2\Rightarrow B< \frac{2}{3}\)
    Mà \(3A=B-\frac{100}{2^{100}}\Rightarrow3A< B< \frac{2}{3}\Rightarrow A< \frac{2}{9}\)

  2. A.2=1-2/2+3/2^2-4/2^3+….+99/2^98-100/2^99
    A.2-A=1-1/2-1/2^2-1/2^3-…-1/2^99-100/2^100
    A=1-(1/2+1/2^2+1/2^3+1/24+…+1/2^99)-100/2^100
    A=1-(1-1/2^99)-100/2^100
    A=1-1+1/2^99-100/2^100
    A=1/2^99-100/2^100
    A=2/2^100-100/2^100
    A=(-98)/2^100
    RỒI BẠN SO SÁNH LÀ SONG

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222-9+11+12:2*14+14 = ? ( )