易知 an = 2n+8
bn=n²-2n
令bn-an=n²-4n-8>0
n>2+2√3或者 n<2-2√3
也就是说
cn=an n<=2+2√3
cn =bn n>2+2√3
an的前n项和为n²+9n
bn的前n项和为 1²+2²+...+n²-2(1+2+...n)
=n(n+1)(2n+1)/6-n(n+1)
=n(n+1)(2n-5)/6
所以
sn = n²+9n n<6
sn=5²+9*5+n(n+1)(2n-5)/6-5*6*5/6
=n(n+1)(2n-5)/6+45 n>=6