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

Toán Lớp 6: Bài 10: Cho A = 1 + 3 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7 a, Tính 3A b, Chứng minh A = (3^8 – 1) : 2 Bài 11: Cho Q = 1 + 3 + 3^2 + …

Toán Lớp 6: Bài 10: Cho A = 1 + 3 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7
a, Tính 3A
b, Chứng minh A = (3^8 – 1) : 2
Bài 11:
Cho Q = 1 + 3 + 3^2 + … + 3^2006
a, tính 3q
b, chứng minh q = ( 3^2007 – 1) : 2

Comments ( 2 )

  1. Lời giải và giải thích chi tiết:
    $\text{Bài 10:}$
    $\text{a, A= 1 + 3 +}$ $3^{2}$ $+ 3^{3}$ $+ 3^{4}$ $+ 3^{5}$ $+ 3^{6}$ $+ 3^{7}$
    $\text{3A= 3 +}$ $3^{2}$ $+ 3^{3}$ $+ 3^{4}$ $+ 3^{5}$ $+ 3^{6}$ $+ 3^{7}$ $+ 3^{8}$
    $\text{b, 3A – A = 3 +}$ $3^{2}$ $+ 3^{3}$ $+ 3^{4}$ $+ 3^{5}$ $+ 3^{6}$ $+ 3^{7}$ $+ 3^{8}$ $- 1 – 3 – 3^{2}$ $-3^{3}$ $-3^{4}$ $- 3^{5}$ $- 3^{6}$ $- 3^{7}$
    $\text{2A = }$ $3^{8}$ $-1^{}$
    $\text{A = }$ $(3^{8}$ $-1)^{}$ $: 2^{}$
    $\text{Bài 11:}$
    $\text{a, Q = 1 + 3 +}$ $3^{2}$ $+…+3 ^{2006}$
    $\text{3Q = 3 +}$ $3^{2}$ $+ 3^{3}$ $+ … +3^{2007}$
    $\text{b, 3Q – Q = }$ $\text{3 +}$ $3^{2}$ $+ 3^{3}$ $+ … +3^{2007}$ $\text{- 1 – 3 -}$ $3^{2}$ $-…-3 ^{2006}$
    $\text{2Q =}$ $3^{2007}$ $- 1^{}$
    $\text{Q =}$ $(3^{2007}$ $- 1)^{} $$: {2}$
     

  2. Giải đáp:
    $\\$
    Bài 10.
    a,
    A =1 + 3 + … + 3^6 + 3^7
    -> 3A = 3 (1+3+…+3^6+3^7)
    -> 3A=3+3^2 + … + 3^7 + 3^8
    Vậy 3A=3+3^2 + … + 3^7 + 3^8
    b,
    Lấy 3A-A ta được :
    -> 3A-A = (3+3^2 + … + 3^7 + 3^8) – (1+3+…+3^6+3^7)
    -> (3-1)A=3^8 – 1
    -> 2A=3^8-1
    -> A=(3^8-1)/2 (đpcm)
    $\\$
    Bài 11.
    a,
    Q = 1 + 3 + … + 3^{2005} + 3^{2006}
    -> 3Q = 3 (1+3+…+3^{2005}+3^{2006})
    -> 3Q=3+3^2+…+3^{2006}+3^{2007}
    Vậy 3Q=3+3^2+…+3^{2006}+3^{2007}
    b,
    Lấy 3Q-Q ta được :
    -> 3Q-Q = (3+3^2+…+3^{2006}+3^{2007}) – (1 + 3 + … + 3^{2005} + 3^{2006})
    -> (3-1)Q=3^{2007}-1
    -> 2Q=3^{2007}-1
    -> Q=(3^{2007}-1)/2 (đpcm)
     

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

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