数列{an}是1,2,4,7,11,......求前n项和Sn

an=a(n-1)+(n-1)
=a(n-2)+(n-1)+(n-2)
=....
=a1+(n-1)+(n-2)+...+1
=1+n(n-1)/2

sn=n+[n(n-1)+(n-1)(n-2)+(n-2)(n-3)+..+3*2+2*1]/2
=n+[n^2-n+(n-1)^-(n-1)+...+3^2-3+2^2-2]/2
=n+[n^2+(n-1)^2+(n-2)^2+...3^2+2^2-(n+2)(n-1)/2]/2
=n+[n*(n+1)*(n+2)/6-1-(n+2)(n-1)/2]/2

这种列叫一阶等差列，由性质，可书：a1=1,a2=a1+1,a3=a1+1+2,then imply:
an= a1+1+2+...+n=a1+(n-1).n/2 = n(n-1)/2+1
求和就是
sn=n+[n(n-1)+(n-1)(n-2)+(n-2)(n-3)+..+3*2+2*1]/2
=n+[n^2-n+(n-1)^-(n-1)+...+3^2-3+2^2-2]/2
=n+[n^2+(n-1)^2+(n-2)^2+...3^2+2^2-(n+2)(n-1)/2]/2
=n+[n*(n+1)*(n+2)/6-1-(n+2)(n-1)/2]/2

差后等差