Question

在数列{an}中a1=2,a(n+1)=an+In(1+1/n),则an=?

Answer 1

a(n+1)=an+ln(1+1/n)=an+ln[(n+1)/n]=an+ln(n+1)-lnn
a(n+1)-an=ln(n+1)-lnn
an-a(n-1)=lnn-ln(n-1)
……
a3-a2=ln3-ln2
a2-a1=ln2-ln1=ln2
连加，有：
an-a1=lnn
an=lnn+a1=2+lnn (n>=2)
n=1也适合
an=2+lnn

Answer 2

无法用初等函数表示

Answer 3

a(n+1)=an+In(1+1/n)
a(n+1)-In(n+1)=an-Inn
a1＝2
a(n+1)-In(n+1)=an-Inn＝。。。＝a1－In1＝2
a(n+1)-In(n+1)=2
a(n+1)＝In(n+1)＋2
an＝Inn＋2

Answer 4

a(n+1)=an+In(1+1/n)
“In”是loge吗？

Answer 5

2+In n