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数学
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在数列{an}中,a1=2,a(n+1)=an+ln(1+1/n)
an为多少
人气:139 ℃ 时间:2020-09-16 23:07:48
解答
A(n+1)=An+ln(1+1/n)
a(n+1)-an=ln(1+1/n)=ln【(n+1)/n】
an=a1+(a2-a1)+(a3-a2)+(a4-a3)+.+(an-an-1)
=2+ln(2/1)+ln(3/2)+ln(4/3)+.+ln(n/n-1)
=2+ln(2/1*3/2*4/3*...*n/n-1)
=2+lnn
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