Môn Toán Lớp 6 tính hợp lý: k) 2^100 – 2^99 – 2^98 – … – 2^2 – 2 – 1 =

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Môn Toán Lớp 6 tính hợp lý: k) 2^100 – 2^99 – 2^98 – … – 2^2 – 2 – 1 = Giúp em bài này với ạ em cần gấp, đừng copy nguồn trên mạng nha. Em xin cảm ơn thầy cô và các bạn nhiều.

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Rose 6 tháng 2022-05-29T03:14:37+00:00 2 Answers 0 views 0

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    0
    2022-05-29T03:15:49+00:00

    `k,2^100-2^99-2^98-. . .-2^2-2-1`

    `=2^100-(2^99+2^98+. . .+2^2+2+1)`

    `=2^100-(1+2+2^2+. . .+2^99)`

    Đặt `A=1+2+2^2+. . .+2^99`

    `2A=2+2^2+2^3+. . .+2^100`

    `2A-A=(2+2^2+. . .+2^100)-(1+2+. . .+2^99)`

    `A=2^100-1`

    `⇒2^100-(1+2+2^2+. . .+2^99)`

    `=2^100-(2^100-1)`

    `=2^100-2^100+1`

    `=0+1`

    `=1`

    0
    2022-05-29T03:16:13+00:00

    $\text{ Đặt A = $2^{100}$ – $2^{99}$ – $2^{98}$ – … – $2^2$ – 2 – 1 }$

    $\text{ = $2^{100}$ – ( $2^{99}$ + $2^{98}$ + … + $2^2$ + 2 + 1 ) }$

    $\text{ = $2^{100}$ – ( 1 + 2 + $2^2$ + … + $2^{98}$ + $2^{99}$ ) }$

    $\text{ Đặt B = 1 + 2 + $2^2$ + … + $2^{98}$ + $2^{99}$ }$

    $\Rightarrow$ $\text{ 2B = 2 + $2^2$ + $2^3$ + … + $2^{99}$ + $2^{100}$ }$

    $\Rightarrow$ $\text{ 2B – B = ( 2 + $2^2$ + $2^3$ + … + $2^{99}$ + $2^{100}$ ) – ( 1 + 2 + $2^2$ + … + $2^{98}$ + $2^{99}$ ) }$

    $\Rightarrow$ $\text{ B = $2^{100}$ – 1 }$

    $\Rightarrow$ $\text{ A = $2^{100}$ – ( $2^{100}$ – 1 ) }$

    $\Rightarrow$ $\text{ A = $2^{100}$ – $2^{100}$ + 1 }$

    $\Rightarrow$ $\text{ A = 0 + 1 }$

    $\Rightarrow$ $\text{ A = 1 }$

      $\text{ Vậy $2^{100}$ – $2^{99}$ – $2^{98}$ – … – $2^2$ – 2 – 1 = 1 }$

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