若数列{an}由a1=2,an+1=an+2n(n≥1),确定,则a100的值为 _.

问题描述:

若数列{an}由a1=2,an+1=an+2n(n≥1),确定,则a100的值为 ______.

∵an+1=an+2n∴an-an-1=2(n-1)=2n-2an-1-an-2=2(n-2)=2n-4…a3-a2=2×2=4=4a2-a1=2=2将上面(n-1)个式子相加可得:an-a1=n×(2n)+{n(0+[-2(n-1)])/2}=n2-n∴a100=1002-100+2=10000-100+2=9902故答案为:9...