（Ⅰ）由an+1=an-1n(n+1)移向得an+1-an=-1n(n+1)=1n+1−1n当n≥2时，an=（an-an-1）+（an-1-an-2）+…+（a2-a1）+a1=（1n−1n−1）+（1n−1−1n−2）+…+（12−11）+2=1n+1．当n=1时，也适合上式，所以数列{an}的通...