`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`
