数列｛an}满足a（n+1)=3an+n（n属于正整数）,是否存在a1,使｛an}成等差数列

a(n+1) = 3a(n) + n, n = 1,2,...
设
a(n+1) + b(n+1) + c = 3[a(n) + bn + c],
则
n = 3bn + 3c - bn - b - c = 2bn + 2c - b,
b = 1/2, c = b/2 = 1/4.
a(n+1) + (n+1)/2 + 1/4 = 3[a(n) + n/2 + 1/4],
设b(n) = a(n) + n/2 + 1/4, n = 1,2,...
则 b(n+1) = 3b(n).
{b(n)}是首项为b(1) = a(1) + 1/2 + 1/4 = a(1) + 3/4,公比为3的等比数列。
b(n) = [a(1)+3/4]3^(n-1), n = 1,2,...
a(n) = b(n) - n/2 - 1/4 = [a(1)+3/4]3^(n-1) - n/2 - 1/4, n = 1,2,...
a(n+1) = [a(1)+3/4]3^n - (n+1)/2 - 1/4
a(n+1) - a(n) = [a(1)+3/4]3^n - (n+1)/2 - 1/4 - {[a(1)+3/4]3^(n-1) - n/2 - 1/4}
= 2[a(1)+3/4]3^(n-1) - 1/2
要使得 a(n+1) - a(n)与n无关，只能a(1)= -3/4.
当a(1) = -3/4时，
a(n) = -n/2 - 1/4. n = 1,2,...
{a(n)}成等差数列。

首先假设存在a1使｛an}成等差数列,则a1+a3=2a2,设公差为d.由｛an}满足a(n+1)=3an+n可知a2=3a1+1；a3=3a2+2=3(a1+d)+2.因此a1+3(a1+d)+2=2(3a1+1)化简后得a1=3d/2.由于公差为d,所以a2=a1+d=5d/2；a3=a2+d=7d/2；a3=a2...