Cho đa thức f(x)=1( x+1 )+( x+1 )( x+2 )+( x+2 )(x+3)×…+( x+49 )( x+50 ). Tính f(1)

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Cho đa thức f(x)=1( x+1 )+( x+1 )( x+2 )+( x+2 )(x+3)×…+( x+49 )( x+50 ). Tính f(1)

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Reese 3 tháng 2021-08-30T21:19:51+00:00 1 Answers 10 views 0

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    2021-08-30T21:21:05+00:00

    `f(x)=1( x+1 )+( x+1 )( x+2 )+( x+2 )(x+3)+…+( x+49 )( x+50)`
    `=> f(1) = 1.(1+1) + (1+1).(1+2) + (1+2).(1+3)+..+(1+49).(1+50)`
    `=> f(1) = 1.2 + 2.3 + 3.4 + … + 50.51`
    `=> 3.f(1) = 1.2.3 + 2.3.3 + 3.4.3 + … + 50.51.3`
    `=> 3.f(1) = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + … + 50.51.(52-49)`
    `=> 3.f(1) = 1.2.3 + 2.3.4 – 1.2.3 + 3.4.5 – 2.3.4 + 50.51.52 – 49.50.51`
    `=> 3.f(1) = 50.51.52`
    `=> f(1) = (50.51.52) :3`
    `=> f(1) = 44200`

    Vậy `f(1) = 44200`

     

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