已知各项均不为0的数列 an 前[n]项和为sn,an+4SnSn-1=0(n≥2),a1=1/2,Bn=4/SnS(n

Bn=4/SnS(n-1)有问题 我觉得 SnS(n-1)是分子 4是分母 Bn=1/[4(4n-2)(4n-6)]=(1/16)[1/(4n-6)-1/(4n-2)] T100=B1+B2+B3+...+B100 =(1/16)[1/(-2)-1/2+1/2-1/6+1/6-1/10+....+1/394-1/398] =(1/16)(-1/2-1/398) =-(1/16)(200/398) =-25/796

Bn=64n²-128n+48的话 T100=B1+B2+B3+...+B100 =64(1²+2²+3²+...+100²)-128*(1+100)*100/2+48*100 =64(1²+2²+3²+...+100²)-128*(1+100)*100/2+48*100 =21654400-646400+4800 =21012800

由题意可知： an=Sn-S(n-1) 则 Sn-S(n-1)+4SnS(n-1)=0 1/S(n-1)-1/Sn+4=0 1/Sn-1/S(n-1)=4 {1/Sn}为等差数列 首项 1/S1=1/a1=2 公差d=4 1/Sn=2+4(n-1)=4n-2 Sn=1/(4n-2) S(n-1)=1/(4n-6) Bn=4/SnS(n-1) =4/{[1/(4n-2)][1/(4n-6)]] =4(4n-2)(4n-6) =64n²-128n+48 T100=B1+B2+B3+...+B100 =64(1²+2²+3²+...+100²)-128*(1+2+3+...+100)+48*100 =64(1²+2²+3²+...+100²)-128*(1+100)*100/2+48*100 =21654400-646400+4800 =21012800 你看看问题在哪里