已知数列{an}满足a(n+1)=an+n,a1=1,则an=

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已知数列{an}满足a(n+1)=an+n,a1=1,则an=已知数列{an}满足a(n+1)=an+n,a1=1,则an=已知数列{an}满足a(n+1)=an+n,a1=1,则an=a(n+1)=

已知数列{an}满足a(n+1)=an+n,a1=1,则an=
已知数列{an}满足a(n+1)=an+n,a1=1,则an=

已知数列{an}满足a(n+1)=an+n,a1=1,则an=
a(n+1)=an+n即a(n+1)-an=n
所以n>=2时有
an-a(n-1)=n-1
.
a2-a1=1
以上n-1个式子相加得到
an-a1=1+2+3+...+（n-1）=n*(n-1)/2
an=n*(n-1)/2+1
n=1时,a1=1*(1-1)/2+1=1,也满足上式,所以
an=n*(n-1)/2+1

对n>1，a(n)=a(n-1)+(n-1)=a(n-2)+(n-2)+(n-1)=a(n-3)+(n-3)+(n-2)+(n-1)=......=a(1)+1+2+3+...+(n-2)+(n-1)=a(1)+n(n-1)/2=1+n(n-1)/2，从而，a(n)=1+n(n-1)/2,n>1.易知，此通项公式对n=1同样成立。

an-a(n-1)=n-1
a(n-1)-a(n-2)=n-2
......
a2-a1=1
an-1=(n-1)+(n-2)+...+2+1=(n-1)n/2
an=1+[(n-1)n/2]

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