b) Chứng tỏ rằng 2(u₁ + u₂ + u₃ + ... + u_n) = n(u₁ + u_n).

b) Ta có: 2(u₁ + u₂ + u₃ + ... + u_n)

= 2[(u₁ + u_n) + (u₂ + u_n-1) + (u₃ + u_n-2) + ... + (u_k + u_n-k+1)]

= 2[(u₁ + u_n)  + (u₁ + u_n) + ... + (u₁ + u_n)]

=  $2.\frac{n}{2}\left(u_1+u_n\right)$ = n(u₁ + u_n) .