设数列an,bn分别满足a1*a2*a3...*an=1*2*3*4...*n,b1+b2+b3+...bn=an^2,n属于N+
问题描述：
设数列an,bn分别满足a1*a2*a3...*an=1*2*3*4...*n,b1+b2+b3+...bn=an^2,n属于N+
a1*a2*a3...*an=1*2*3*4...*n,b1+b2+b3+...bn=an^2,n属于N+
1)求数列an和bn的通项公式
答
a1*a2*a3...*an*a(n+1)=1*2*3*4...*n*(n+1)
a1*a2*a3...*an=1*2*3*4...*n
两式相除
=> a1=1 ,a(n+1) = n+1 => an=n
b1+b2+b3+...bn=an^2=n^2
b1+b2+b3+...bn+b(n+1)=a(n+1)^2=(n+1)^2
两式相减
=> b1=1 ,b(n+1) = (n+1)^2 - n^2 = 2n+1
=> bn=2n-1