Question

a1=2，a(n+1)=an+n+1，求通项an=

Answer 1

a2=a1+1+1
a3=a2+2+1
a4=a3+3+1
……
a(n-1)=a(n-2)+n-2+1
an=a(n-1)+n-1+1
左右相加得
an=a1+1+2+3+...+n-1+n-1
=2+nx(n+1)/2 -1
=n^2/2+n/2+1

Answer 2

a(n+1)=an+n+1
an=a(n-1)+n-1+1
a(n-1)=a(n-2)+n-2+1
...
a3=a2+2+1
a2=a1+1+1
将等式左右分别相加,消项得:an=a1+1+2+3+..+n-1+n-1
an=n(n-1)/2+n+1