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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1. `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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