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

Toán Lớp 6: CMR:A= 1 phần 3+ 2 phần3^2+3 phần 3^3+4 phần 3^4+5 phần 3^5+…+102 phần 3^102< 3 phần 4

Tháng Một 6, 2023 Toán học 1 Comment 0 views

Toán Lớp 6: CMR:A= 1 phần 3+ 2 phần3^2+3 phần 3^3+4 phần 3^4+5 phần 3^5+…+102 phần 3^102< 3 phần 4

Comments ( 1 )

  1. truongthanhhung
    truongthanhhung
    6 Tháng Một, 2023 at 7:59 chiều
    Reply
    A = $\frac{1}{3}$ + $\frac{2}{3^{2}}$ + $\frac{3}{3^{3}}$ + … + $\frac{102}{3^{102}}$
    $\frac{1}{3}$ A = $\frac{1}{3^{2}}$ + $\frac{2}{3^{3}}$ + $\frac{3}{3^{4}}$ + … + $\frac{102}{3^{103}}$
    A – $\frac{1}{3}$ A = ($\frac{1}{3}$ + $\frac{2}{3^{2}}$ + $\frac{3}{3^{3}}$ + … + $\frac{102}{3^{102}}$) –  ($\frac{1}{3^{2}}$ + $\frac{2}{3^{3}}$ + $\frac{3}{3^{4}}$ + … + $\frac{102}{3^{103}}$)
    $\frac{2}{3}$ A = $\frac{1}{3}$ + $\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{102}}$ – $\frac{102}{3^{103}}$
    A = ($\frac{1}{3}$ + $\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{102}}$ – $\frac{102}{3^{103}}$) · $\frac{3}{2}$
    Ta có: A = ($\frac{1}{3}$ + $\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{102}}$ – $\frac{102}{3^{103}}$) · $\frac{3}{2}$ < ($\frac{1}{3}$ + $\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{102}}$) · $\frac{3}{2}$
    Đặt B = ($\frac{1}{3}$ + $\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{102}}$) · $\frac{3}{2}$
    Ta có:
    B = ($\frac{1}{3}$ + $\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{102}}$) · $\frac{3}{2}$
    $\frac{1}{3}$ B = ($\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{103}}$) · $\frac{3}{2}$
    B – $\frac{1}{3}$ B = ($\frac{1}{3}$ + $\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{102}}$) · $\frac{3}{2}$ – ($\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{103}}$) · $\frac{3}{2}$
    $\frac{2}{3}$ B = $\frac{3}{2}$ · [($\frac{1}{3}$ + $\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{102}}$) – ($\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + … + $\frac{1}{3^{103}}$)]
    B = $\frac{9}{4}$ · [$\frac{1}{3}$ – $\frac{1}{3^{103}}$]
    B = $\frac{9}{4}$ · $\frac{3^{102} – 1}{3^{103}}$
    B = $\frac{1}{4}$ · $\frac{3^{102} – 1}{3^{101}}$ < $\frac{1}{4}$ · $\frac{3^{102}}{3^{101}}$
    Đặt C = $\frac{1}{4}$ · $\frac{3^{102}}{3^{101}}$
    Ta có:
    C = $\frac{1}{4}$ · 3
    C = $\frac{3}{4}$
    ⇒ A < B < C = $\frac{3}{4}$
    Vậy A < $\frac{3}{4}$

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

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