已知数列{an}满足：a1=3，an=3an−1（n≥2）．

（1）当n=1时，a₁=3，
假设当n=k，a_k=4m_k+3，m_k∈N．
则当n=k+1时，ak+1＝3ak
=34mk+3
=(4−1)4mk+3
=
C04mk+344mk+3+
C14mk+344mk+2(−1)+…+
C4mk+24mk+3•4•(−1)4mk+2+
C4mk+34mk+3(−1)4mk+3
=4T-1=4（T-1）+3．其中T=
C04mk+344mk+2+
C′14mk+344mk+1(−1)+…+
C4mk+24mk+3(−1)4mk+2∈N^*．
∴∃m_k+1=∈N使a_k+1=4m_k+1+3，
∴当n=k+1时，结论也成立．
∴∀n∈N^*，∃m_n∈N，使a_n=4m_n+3．
（2）an+1＝3an=34mn+3=81mn×27，
故a₂₀₁₃的末位数字是7．

点评：
本题考点： 数学归纳法．

考点点评： 本题考查了数学归纳法和二项式定理的性质，属于难题．