已知数列｛an｝的首项a1=a（a为常数）,an=2an-1+n^2-4n+2（n大于等于2）

a(n+1) = 2a(n) + (n+1)^2 - 4(n+1) + 2 = 2a(n) + n^2 + 2n + 1 - 4n - 4 + 2 = 2a(n) + n^2 -2n - 1
= 2a(n) + 2(n-1)^2 - n^2 + 2n - 3
= 2a(n) + 2(n-1)^2 - n^2 + 4(n-1) - 2n + 1
= 2a(n) + 2(n-1)^2 + 4(n-1) - n^2 - 2n + 2 - 1,
a(n+1) + n^2 + 2n + 1 = 2a(n) + 2(n-1)^2 + 4(n-1) + 2 = 2[a(n) + (n-1)^2 + 2(n-1) + 1],
{a(n) + (n-1)^2 + 2(n-1) + 1}是首项为a(1) + 1 = a+1,公比为2的等比数列.
a(n) + (n-1)^2 + 2(n-1)+ 1 = (a+1)2^(n-1) = a(n) + n^2,
a(n) = (a+1)2^(n-1) - n^2.
a(n+1) = (a+1)2^n - (n+1)^2,
a(n+1) - a(n) = 2(a+1)2^(n-1) - (a+1)2^(n-1) - (n+1)^2 + n^2 = (a+1)2^(n-1) - (2n+1),与n相关,不是常数,因此,{a(n)}不是等差数列.
b(1) = b.
n>=2时,
b(n) = a(n) + n^2 = (a+1)2^(n-1),
若a+1=b,则{b(n) = (a+1)2^(n-1)}是首项为b=a+1,公比为2的等比数列.
是s(n) = b(1) + b(2) + ...+ b(n)吧?
n=1 时,s(1) = b(1) = b.
n>=2时,
s(n) = b(1) + b(2)+b(3)+...+b(n) = b + (a+1)[2+2^2+...+2^(n-1)]
= b + 2(a+1)[1 + 2 + ...+ 2^(n-2)]
= b + 2(a+1)[2^(n-1) - 1]/(2-1)
= b + 2(a+1)[2^(n-1) - 1]
= b - 2(a+1) + 2(a+1)*2^(n-1),
因此,总有,
s(n) = b - 2(a+1) + 2(a+1)*2^(n-1).
若s(n) = bq^(n-1),则
b - 2(a+1) + 2(a+1)2^(2-1) = s(2) = bq = b + 2(a+1),2(a+1) = b(q-1),
s(n) = b - 2(a+1) + 2(a+1)*2^(n-1) = b - b(q-1) + b(q-1)2^(n-1) = b(2-q) + b(q-1)2^(n-1) = bq^(n-1),
s(3) = b(2-q) + b(q-1)2^2 = bq^2 = 2b - bq + 4bq - 4b = 3bq - 2b,
0 = bq^2 - 3bq + 2b = b[q^2 - 3q + 2] = b(q-2)(q-1),
b=0与题意不符.因此,b不为0.
q=1时,2(a+1) = b(q-1) = 0,a = -1,b为任意非零常数.{s(n) = b}是首项为s(1)=b,公比为1的等比数列,符合题意.
q=2时,2(a+1) = b(q-1) = b.{s(n) = b2^(n-1)}是首项为s(1)=b,公比为2的等比数列,符合题意.
综合,有,
a=-1,b为任意非零常数时,{s(n)=b }是首项为s(1)=b,公比为1的等比数列.
a不为-1,b=2(a+1)时,{s(n) = b2^(n-1)}是首项为s(1)=b,公比为2的等比数列.