已知数列an的首项a1=2a+1(a是常数,且a≠-1),an=2a(n-1)+n²-4n+2(n≠-1）

1,n>=2时,bn=an+n^2=2a(n-1)+2(n^2-2n+1)=2a(n-1)+2(n-1)^2=2[a(n-1)+(n-1)^2]=2bn
所以数列{bn}从第下项起是公比为2的等比数列.
2,S1=b1=a S2=B1+b2=a+a2+2^2=a+2a1+1^2-4*1+2=a+2a+1-1=3a
S3=S2+b3=S2+2b2=S2+2(a2+2^2)=S2+2a2+8=3a+2(2a1+1^2-4*1+2)+8=13a+8
(3a)^2=a(13a+8) a=-2或a=0(舍去)
3,an=2a(n-1)+n²-4n+2两边同加n^2并整理得：an+n^2=2[a(n-1)+(n-1)^2]
所以数列{an+n^2}是首项为2a+2、公比为2的等比数列,通项为an+n^2=(a+1)*2^n
an=(a+1)*2^n-n^2
设f(x)=(a+1)*2^x-x^2(x>=1) f'(x)=(a+1)ln2*2^x-2x f''(x)=(a+1)(ln2)^2*2^x-2>2^x-2>0
所以f'(x)在x>=1时是增函数,即x>=1时,f'(x)>=f'(1)=(a+1)ln2*2-2>0
所以f(x)在x>=1时是增函数,即数列{an}是递增数列,a1=2a+1是an的最小值.