Lời giải:

a) Ta có: u₂ = u₁ + d; ...; u_{n – 1} = u₁ + (n – 1 – 1)d = u₁ + (n – 2)d; u_n = u₁ + (n – 1)d.

S_n = u₁ + u₂ + ... + u_{n – 1} + u_n

= u₁ + (u₁ + d) + ... + [u₁ + (n – 2)d] + [u₁ + (n – 1)d]

b) S_n = u_n + u_{n – 1} + ... + u₂ + u₁

= [u₁ + (n – 1)d] + [u₁ + (n – 2)d] + ... + (u₁ + d) + u₁

c) Ta có:

S_n + S_n = {u₁ + (u₁ + d) + ... + [u₁ + (n – 2)d] + [u₁ + (n – 1)d]} + {[u₁ + (n – 1)d] + [u₁ + (n – 2)d] + ... + (u₁ + d) + u₁}

⇔ 2S_n = {u₁ + [u₁ + (n – 1)d]} + {(u₁ + d) + [u₁ + (n – 2)d]} + ... + {[u₁ + (n – 2)d] + (u₁ + d)} + {[u₁ + (n – 1)d] + u₁}

⇔ 2S_n = [2u₁ + (n – 1)d] + [2u₁ + (n – 1)d] + ... + [2u₁ + (n – 1)d] + [2u₁ + (n – 1)d]

⇔ 2S­_n = n . [2u₁ + (n – 1)d]

⇔ S_n = \(\frac{n}{2}\)[2u₁ + (n – 1)d] .