Question

CMR: A = 1³+ 2³+ 3³ + ...+ 100³ chia heets cho B = 1 +2+3+...+100

Answer 1

Ta có: B=1+2+3+...+100

=(1+100)+(2+99)+...+(50+51)

\(=101\cdot50\)

Ta có: \(A=1^3+2^3+3^3+...+100^3\)

\(=\left(1^3+100^3\right)+\left(2^3+99^3\right)+...+\left(50^3+51^3\right)\)

\(=\left(1+100\right)\cdot\left(1-100+100^2\right)+\left(2+99\right)\left(4-198+99^2\right)+...+\left(50+51\right)\left(2500+50\cdot51+51^2\right)\)

\(=101\cdot\left(1-100+100^2+4-198+99^2+...+50^2-50\cdot51+51^2\right)⋮101\)

Ta có: \(A=1^3+2^3+3^3+...+100^3\)

\(=\left(1^3+99^3\right)+\left(2^3+98^3\right)+...50^3+100^3\)

\(=\left(1+99\right)\left(1-99+99^2\right)+\left(2+98\right)\cdot\left(4-196+98^2\right)+...+50^3+50^3\cdot2^3⋮50\)

mà (50,101)=1

nên \(A⋮50\cdot101=B\)

hay \(A⋮B\)(đpcm)