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

Toán Lớp 6: giúp e vs 3+3 ²+3 ³+…+3^2020 CMR: tổng trên chia hết cho 120

Toán Lớp 6: giúp e vs
3+3 ²+3 ³+…+3^2020
CMR: tổng trên chia hết cho 120

Comments ( 2 )

  1. $3 + 3^{2} + 3^{3} + … 3^{2020}$
    $= (3 + 3^{2} + 3^{3} + 3^{4}) + (3^{5} + 3^{6} + 3^{7} + 3^{8}) + … + (3^{2017} + 3^{2018} + 3^{2019} + 3^{2020})$
    $= 120 + 3^{4} . (3 + 3^{2} + 3^{3} + 3^{4}) + … + 3^{2016} . (3 + 3^{2} + 3^{3} + 3^{4})$
    $= 120 + 3^4 . 120 + … + 3^{2016} . 120$
    $= 120 . (1 + 3^{4} + … + 3^{2016})$
    → \text{Tổng trên chia hết cho 120.}

  2. $\text{ 3 + $3^2$ + $3^3$ + … + $3^{2020}$ }$
    $\text{ = ( 3 + $3^2$ + $3^3$ + $3^4$ ) + … + ( $3^{2017}$ + $3^{2018}$ + $3^{2019}$ + $3^{2020}$ ) }$
    $\text{ = ( 3 + $3^2$ + $3^3$ + $3^4$ ) + … + $3^{2016}$. ( 3 + $3^2$ + $3^3$ + $3^4$ ) }$
    $\text{ = 120 + … + $3^{2016}$. 120 }$
    $\text{ = 120. ( 1 + $3^{2016}$ )      $\vdots$ 120  ( đpcm ) }$
     

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

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