已知数列{an}满足a1=1,an+1=an+2n,则a10=______.
∵数列{an}满足a1=1,an+1=an+2n,
∴an=a1+(a2-a1)+…+(an-an-1)=1+21+22+…+2n-1=
=2n-1.(n∈N*).2n-1 2-1
∴a10=210-1=1023.
故答案为:1023.
已知数列{an}满足a1=1,an+1=an+2n,则a10=______.
∵数列{an}满足a1=1,an+1=an+2n,
∴an=a1+(a2-a1)+…+(an-an-1)=1+21+22+…+2n-1=
=2n-1.(n∈N*).2n-1 2-1
∴a10=210-1=1023.
故答案为:1023.