填空题在数列{an}中,a1=1,且对于任意正整数n,都有an+1=an+n,则a100=________.
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4951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)=1+1+2+…+99=4951.解答:∵a1=1,an+1=an+n,∴a2-a1=1,a3-a2=2,…,a100-a99=99,∴a100=a1+(a2-a1)+(a3-a2)+…+(a100-a99)=1+1+2+…+99=4951.