一道高一数学题

显示全部楼层 · 2013-7-27 17:28:34

a5=9，a2+a6=14a1+4d=9,a1+3d=7a1=1,d=2an=1+2(n-1)=2n-1bn=a(n+1)=2n+1b(n+1)=2n+3Tn=1/（bn*b（n+1）=1/(2n+1)(2n+3)=1/2*[1/(2n+1)]-[1/(2n+3)]Tn=T1+T2+.....+Tn=1/2{(1/3)-(1/5)+(1/5)-(1/7)+...+1/(2n+1)]-[1/(2n+3)}=1/2[(1/3)-1/(2n+3)]=1/(6n+9)...

千问 · 2013-7-27 17:28:34

a2+a6=2a4=14a4=7a5=9公差d=2, a1=a5-4d=9-8=1所以an=2n-1bn=a(n+1)=2n+11/(bn*b(n+1))=1/2(1/bn-1/b(n+1)通过裂项求和，得到Tn=1/2(1/b1-1/b(n+1))=1/2(1/3-1/(2n+3))=n/(3(2n+3))...

千问 · 2013-7-27 17:28:34

由a5=9，a2+a6=14得an=2n-1，所以bn=2n+1，所以1/（bn*b（n+1））=1/2【1/bn-1/b（n+1）】，所以Tn=1/2【1/b1-1/bn】...

千问 · 2013-7-27 17:28:34

An=2n-1，所以1/bn*bn+1=1/（2n+1）*（2n+3）=0.5(1/（2n+1）-1/（2n+3）)所以Tn=0.5(1/3-1/5+1/5-1/7+...+1/（2n+1）-1/（2n+3）)=0.5*(1/3-1/(2n+3))...