问题
填空题
已知数列{an}满足an+1=an+n,a1=1,则an=______.
答案
∵an+1=an+n,a1=1,
∴a2-a1=1
a3-a2=2
…
an-an-1=n-1
以上n-1个式子相加可得,an-a1=1+2+…+n-1=
×(n-1)=1+n-1 2 n2-n 2
∴an=
n2-1 2
n+11 2
故答案为:an=
n2-1 2
n+11 2
已知数列{an}满足an+1=an+n,a1=1,则an=______.
∵an+1=an+n,a1=1,
∴a2-a1=1
a3-a2=2
…
an-an-1=n-1
以上n-1个式子相加可得,an-a1=1+2+…+n-1=
×(n-1)=1+n-1 2 n2-n 2
∴an=
n2-1 2
n+11 2
故答案为:an=
n2-1 2
n+11 2