Question

已知数列｛an｝的前n项和是Sn,a1=2且Sn*S(n-1)+1/2an=0

,(1)求Sn (2)1/S1S2-1/S2S3+1/S3S4-1/S4S5+…+1/S99S100-1/S100S101

Answer 1

1.
Sn*S(n-1)+(1/2)an=0
Sn*S(n-1)+(1/2)[Sn-S(n-1)]=0
1+(1/2)[1/S(n-1)-1/Sn]=0
1/Sn-1/S(n-1)=2
1/Sn=1/S1-2(n-1)=1/2-2n+2=5/2-2n
Sn=2/(5-4n)

2.
1/S1S2-1/S2S3+1/S3S4-1/S4S5+…+1/S99S100-1/S100S101
=(1/S2)(1/S1-1/S3)+(1/S4)(S3-1/S5)+…+(1/S100)(1/S99-1/S101)
=(1/S2)(5/2-2-5/2+2*3)+(1/S4)(5/2-2*3-5/2+2*5)+…+(1/S100)(5/2-2*99-5/2+2*101)
=4/S2+4/S4+…+4/S100
=4[5/2-2*2]+4[5/2-2*4]+…+4[5/2-2*100]
=4(5/2)*50-4[2*2+2*4+…+2*100]
=500-16[1+2+3+…+50]
=500-16*51*50/2
=-19900