高中数学题急~~在线等

那个是N+1项=N项+ln(1+1/n)吧
a n+1-an=ln(n+1/n) =ln(n+1)-ln n
所以 a2-a1=ln2-ln1
a3-a2=ln3-l2
......
an-a n-1=ln(n+1)-ln n
再用累加法 可得an=ln n-ln1=ln n (n>=2) 所以AN的通项为 n=1时 an =2
n>=2时 an =ln n

an - a(n-1)= ln n/(n-1)
a(n-1) - a(n-2) = ln (n-1)/(n-2)
……
a2 - a1 = ln2/1
左右相加
an-a(n-1)+a(n-1) -……-a1 = ln (1+n)/n + ln n/(n-1)+……ln 2/1 = ln (1+n)/n * … 2/1 = ln n
an = 2+ln n

A（n＋1）＝An＋ln（1＋1/n)
an+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

叠加法，an=a(n-1)+ln[n/(n-1)]
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a2=a1+ln2
an=lnn+a1=lnn+2

a(n+1)=an+ln(1+1/n)=an+ln[(n+1)/n]=an+ln(n+1)-lnn
a(n+1)-ln(n+1)=an-lnn

an=lnn+k
a1=2=k

an=lnn+2