Question

数列{an}的前n项和为Sn，已知a1=1/2,Sn=n^2an-n(n-1) (1)证明：数列{[﹙n+1)Sn]/n}是等差数列

步骤详细点
设bn=Sn/n,求满足bn>2012/2013的最小正整数n.

Answer 1

追问

设bn=Sn/n,求满足bn>2012/2013的最小正整数n.

追答

Answer 2

s[n+1]-s[n]得:

a[n+1]=a[n+1]*(n+1)^2-a[n]*n^2-2n
令b[n]=1-a[n]
1-b[n]=(1-b[n])*(n+1)^2-(1-b[n])*n^2-2n
b[n+1]*(n+2)n=b[n]*n^2
b[n+1]=b[n]*n/(n+2)=b[n-1]*n*(n-1)/((n+2)*(n+1))=...=b1*n*(n-1)*...*1/((n+2)*(n+1)*...*3)

b1=1/2
b[n+1]=1/2*2/((n+2)*(n+1))=1/((n+1)*(n+2))=1/(n+1)-1/(n+2)

b[n]=1/n-1/(n+1)
a[n]=1-1/n+1/(n+1)
s[n]=n-1+1/(n+1)=n^2/(n+1)
﹙n+1)Sn]/n=n

故，为等差数列。

本题难点在于b[n]的选取，一旦选好，则本题难度大减，关于b[n]的选取，一般根据经验或者待定系数法或者通过前几项的规律总结，总之，多做题，熟能生巧。