已知数列{an},a1=2,an=an-1+log2(1+2/n)（n≥2）求anan=a（n-1）+log2(1+2/n)log2(1+2/n)的第一个2是底下的小2,

Question

已知数列{an},a1=2,an=an-1+log2(1+2/n)（n≥2）求an
an=a（n-1）+log2(1+2/n)
log2(1+2/n)的第一个2是底下的小2,

Answer 1

由已知：an=a(n-1)+log2(1+2/n),可得an-a(n-1)=log2((n+2)/n))=log2(n+2)-log2n
an=(an-a(n-1))+(a(n-1)-a(n-2))+(a(n-2)-a(n-3))+……+(a2-a1)+a1=[log2(n+2)-log2n]+[log2(n+1)-log(n-1)]+[log2n-log2(n-2)]+……+(log24-log22)+2=log2(n+2)+log2(n+1)-log23-log22+2
=log2[3(n+2)(n+1)]+1