Question

Cho cấp số cộng (U_n). Tính

a) \(S=\frac{1}{U_1U_2}+\frac{1}{U_2U_3}+...+\frac{1}{U_{n-1}U_n}\) theo d , U₁ , U_n

b) \(S=U_1^2+U_2^2+...+U_n^2\) theo d , U₁ , U_n

Answer 1

S= u₁.u₁ + u₂.u₂+...+u_n.u_n 

S = u₁.(u₂ - d) + u₂.(u₃ - d)+...+u_n(u_n+1 - d)

S = u₁.u₂ + u₂.u₃ +...+u_n.u_n+1-d(u₁+u₂+...+u_n)

Đặt A = u₂.u₃ + u₃.u₄+...+u_n.u_n+1

3d.A = u₂.u₃.(u₄-u₁) + u₃.u₄.(u₅-u₂)+...+u_n.u_n+1.(u_n+2-u_n-1) 

3d.A = u₂.u₃.u₄ - u₁.u₂.u₃ + u₃.u₄.u₅ - u₂.u₃.u₄+...+u_n.u_n+1.u_n+2 - u_n-1.u_n.u_n+1

3d.A = u_n.u_n+1.u_n+2 - u₁.u₂.u₃

3d.A = (u₁ + d.n - d)(u₁ + d.n)(u₁ + d.n + d) - u₁.(u₁+d).(u₁+2.d) 

A = [(u₁ + d.n - d)(u₁ + d.n)(u₁ + d.n + d) - u₁.(u₁+d).(u₁+2.d)]/(3.d) 

S = A + u₁.(u₁ + d) + d[2.u₁+(n-1).d].n/2