若数列｛an｝前n项和的公式Sn＝n²＋4n－1,则a6＋a7＋a8＝

a(1)=s(1)=1+4-1=4,
a(n+1)=s(n+1)-s(n) = (2n+1) + 4 = 2n + 5,
n>=2时,a(n) = 2(n-1) + 5,
a(6) + a(7) + a(8) = 2[5+6+7] + 5*3 = 36 + 15 = 51还有1个做法：a(6)+a(7)+a(8) = a(1)+a(2)+...+a(8) - [a(1)+a(2)+...+a(5)] = s(8) - s(5) = 8^2 + 4*8 - 5^2 - 4*5 = 64 + 32 - 25 - 20 = 96 - 45 = 51多谢楼主的及时采纳。。很高兴有机会为楼主效劳~~