Tính : `A = 2(3^2 + 1)(3^4 + 1)(3^8 +1)….(3^{64} + 1)`

Question

Tính :
`A = 2(3^2 + 1)(3^4 + 1)(3^8 +1)….(3^{64} + 1)`

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Acacia 4 years 2020-10-13T20:05:44+00:00 2 Answers 101 views 0

Answers ( )

  1. Tryphena
    0
    2020-10-13T20:06:55+00:00
    Reply

    Đáp án:

     

    Giải thích các bước giải:

     `A=2(3^2+1)(3^4+1)(3^8+1)…(3^64+1)`

    `=>4A=(3+1)(3-1)(3^2+1)(3^4+1)(3^8+1)…(3^64+1)`

    `=>4A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)…(3^64+1)`

    `=>4A=(3^4-1)(3^4+1)(3^8+1)…(3^64+1)`

    `=>4A=(3^8-1)(3^8+1)..(3^64+1)`

    `=>4A=(3^16-1)….(3^64+1)`

    `=>4A=(3^128-1)`

    `=>A=(3^128-1)/4`

  2. King
    0
    2020-10-13T20:07:05+00:00
    Reply

    Đáp án:

    $A = \dfrac{3^{128} -1}{4}$

    Giải thích các bước giải:

    $A= 2(3^2 + 1)(3^4 + 1)(3^8 +1)\dots(3^{64} + 1)$

    $\to A = (3^1 – 1)(3^2 + 1)(3^4 + 1)(3^8 +1)\dots(3^{64} + 1)$

    $\to 4A = (3^1 – 1)(3^1 + 1)(3^2 + 1)(3^4 + 1)(3^8 +1)\cdots(3^{64} + 1)$

    $\to 4A = (3^2 – 1)(3^2 + 1)(3^4 + 1)(3^8 +1)\dots(3^{64} + 1)$

    $\to 4A = (3^4 – 1)(3^4 + 1)(3^8 +1)\dots(3^{64} + 1)$

    $\to 4A = (3^8 -1)(3^8 +1)\dots(3^{64} + 1)$

    $\to 4A = (3^{64} -1)(3^{64} -1)$

    $\to 4A = 3^{128} -1$

    $\to A = \dfrac{3^{128} -1}{4}$

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