数列{an}满足a1=0,an+1=an+n那么a100的值是________.
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4950
解析分析:由题意知a2-a1=1,a3-a2=2,…,a100-a99=99,所以a100=a1+(a2-a1)+(a3-a2)+…+(a100-a99)=0+1+2+…+99=4950.
解答:∵a1=0,an+1=an+n,∴a2-a1=1,a3-a2=2,…,a100-a99=99,∴a100=a1+(a2-a1)+(a3-a2)+…+(a100-a99)=0+1+2+…+99==4950.