设数列{an},a1=2,a(n+1)=an+In·(1+1/n),求an

问题描述:

设数列{an},a1=2,a(n+1)=an+In·(1+1/n),求an

由a(n+1)=an+In(1+1/n)得:an-a(n-1)=ln[1+1/(n-1)]a(n-1)-a(n-2)=ln[1+1/(n-2)]……a2-a1=ln(1+1/1)把上面一串式子加起来,有:an-a1=ln[1+1/(n-1)]+ln[1+1/(n-2)]+……ln(1+1/1)看这个等式的右边关于对数的...