已知数列{an}中，a1=20，an+1=an+2n-1，n∈N*，则数列{an}的通项公式an=______．
因为数列{an}中，a1=20，an+1=an+2n-1，n∈N*，
所以a2=a1+1，
a3=a2+3，
a4=a3+5，
…
an=an-1+2n-3；
上式累加可得：
an=a1+1+3+5+…+（2n-3）=20+n-1+
× 2=n2-2n+21．(n-1)(n-2) 2
故答案为：n2-2n+21．
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