设等差数列an的前n项和为Sn,已知bn=1/Sn,且a3*b3=1/2,S3+S5=21,求bn,bn前n项和Tn._数学解答

设等差数列an的前n项和为Sn,已知bn=1/Sn,且a3*b3=1/2,S3+S5=21,求bn,bn前n项和Tn.

人气：236 ℃　时间：2019-12-12 23:32:57

解答

S3=3a1+3*2*d/2=3a1+3ds5=5a1+5*4*d/2=5a1+10da3*b3=(a1+2d)/S3=(a1+2d)/(3a1+3d)S3+S5=3a1+3d+5a1+10d=8a1+13d(a1+2d)/(3a1+3d)=1/28a1+13d=21解得,a1=1,d=1Sn=na1+n(n-1)d/2=n(n+1)/2bn=1/[n(n+1)/2]=2/[n(n+1)]Tn...