Question

已知等差数列{An}的前n项和为Sn若S10=100.S100=10.求S110

Answer 1

1．由等差数列的前n项和的公式特点,

可设Sn=An²+Bn,

则S10=10²A+10B=100,S100=100²A+100B=10,

两式相减得,(100²-10²)A+(100-10)B= -(100-10),

∴110A+B= -1,

∴S110=110²A+110B=110(110A+B)= -110..

2．设等比数列的公比为q,(q≠0)

则由题意,S4=2S2,

即S4-S2=S2,

∴q²S2=S2,

(q²-1)(a1+a1q)=0

a1(q+1)(q-1)(1+q)=0

∵a1≠0,

∴q=±1,

当q=1时,an=a3=2,

当q= -1时,an=a3q^(n-3)=2×(-1)^(n-3)=2×(-1)^(n-1).