Σai
=Σ[a1+10(i-1)]
=Σa1+10Σ(i-1)
=na1+10(Σi-Σ1)
=na1+10[n(n+1)/2-n]
=na1+10n[(n+1)/2-1]
=na1+10n[(n-1)/2]
=na1+5n(n-1)

Σi=n(n+1)/2之證明
Σi=1+2+···+n (1)
Σi=n+(n-1)+···+1 (2)
(1)+(2)→2Σi=(n+1)+(n+1)+···+(n+1)=n(n+1)
∴Σi=n(n+1)/2