an+1-an=log2 n+1/n,a1=1,求an
问题描述：
an+1-an=log2 n+1/n,a1=1,求an
答
an+1-an=log2 n+1/n 应该是an+1-an=log2（ n+1/n）=log2((n+1)/n)n=1时,a2-a1=log2(2/1)①n>=2时,an-an-1=log2(n/(n-1))②依次写下去...直到：a2-a1=log2(2/1)将这些式子相加消去相同的项得：an-a1=log2(n/(n-1))+.....