设数列{an}中,a1=2,a(n+1)=an+n+1,求an

a(n+1)=a(n)+n+1,
a(n)=a(n-1)+(n-1)+1,
...
a(2)=a(1)+1+1,
等号两边求和.有,
a(n+1)+a(n)+...+a(2)=a(n)+...+a(2)+a(1)+[1+2+...+n]+n,
a(n+1)=a(1)+n(n+1)/2+n=n(n+1)/2+n+2=(n+1-1)(n+1)/2+(n+1)+1,
a(n)=(n-1)n/2+n+1,n=1,2,...
是这样吗? 和答案不同哈...