在数列{an}中,a1=1,3an+1=(1+1⼀n)2an(n∈N*),则数列{an}的通项公式为?

2025-04-28 17:43:26
推荐回答(2个)
回答1:

解答详见下图

回答2:

∵a1=1;3a(n+1)=(1+1/n)²an,
∴3a2=(1+1/1)²×1=(2)²,
a2= [2]²/3;
3a3=(1+1/2)²×[2]²/3=[3/2×2]²/3,
a3= [3/2×2]²/3²
3a4=(1+1/3)²×[3/2×2]²/3²=[4/3×3/2×2]²/3²
a4=[4/3×3/2×2]²/3³
......
an=[n/(n-1)×(n-1)/(n-2)×......×4/3×3/2×2]²/3^(n-1)
=n²/3^(n-1)
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A选项正确