Question

若数列｛an｝前n项和的公式Sn＝n²＋4n－1,则a6＋a7＋a8＝

Answer 1

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