Question

CMR: A CHIA HET CHO B

a,\(A=1^3+2^3+3^3+.....+99+100\)

\(B=1+2+3+...+99+100\)

b,\(A=1^3+2^3+3^3+....+98^3+99^3\)

\(B=1+2+3+...+98+99\)

Ban nao co long tot thi giup minh voi a ,minh can gap!!!!

Answer 1

\(a,\)Biết \(B=\frac{100.101}{2}=50.101\)

\(A=1^3+2^3+3^3+...+99^3+100^3\)

Xét \(A=\left(1^3+100^3\right)+\left(2^3+99^3\right)+...+\left(49^3+52^3\right)+\left(50^3+51^3\right)\)

\(\Rightarrow A=101.\left(1+100+100^2\right)+101.\left(2^2+2.99+99^2\right)+...+101\left(50^2+50.51+51^2\right)\)

\(\Rightarrow A=101\left(1+100+100^2+2^2+2.99+99^2+...+50^2+50.51+51^2\right)⋮101\)

Xét\(A=\left(1^3+99^3\right)+\left(2^3+98^3\right)+...+\left(49^3+51^3\right)+50^3\)

\(\Rightarrow A=100\left(1^2+1.99+99^2\right)+100\left(2^2+2.98+98^2\right)+...+100\left(49^2+49.51+51^2\right)+100.50.25⋮50\)

Vậy \(A⋮101.50=5050=B\)

Làm tương tự với câu b