一道关于数列的填空已知{an},{bn}为等差数列,Sn,Tn分别为其前n项和,若Sn/Tn=(2n+3)/(n+1)
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/11 21:15:37
一道关于数列的填空已知{an},{bn}为等差数列,Sn,Tn分别为其前n项和,若Sn/Tn=(2n+3)/(n+1)
一道关于数列的填空
已知{an},{bn}为等差数列,Sn,Tn分别为其前n项和,若Sn/Tn=(2n+3)/(n+1)
一道关于数列的填空已知{an},{bn}为等差数列,Sn,Tn分别为其前n项和,若Sn/Tn=(2n+3)/(n+1)
a3/b3
=2a3/2b3
=(a1+a5)/(b1+b5)
=[(a1+a5)*5/2]/[(b1+b5)*5/2]
=S5/T5
=(2*5+3)/(5+1)
=13/6
Sn/Tn=[na1+n(n-1)d1/2]/[nb1+n(n-1)d2/2]
=[2a1+(n-1)d1]/[2b1+(n-1)d2]
=(d1n+2a1-d1)/(d2n+2b1-d2)
=(2n+3)/(n+1)
令d1=2t,则2a1-d1=3t d2=t 2b1-2d=t
解得a1=(5/2)t b1=(3/2)t
a3=a1+2d1...
全部展开
Sn/Tn=[na1+n(n-1)d1/2]/[nb1+n(n-1)d2/2]
=[2a1+(n-1)d1]/[2b1+(n-1)d2]
=(d1n+2a1-d1)/(d2n+2b1-d2)
=(2n+3)/(n+1)
令d1=2t,则2a1-d1=3t d2=t 2b1-2d=t
解得a1=(5/2)t b1=(3/2)t
a3=a1+2d1=(5/2)t+4t=(13/2)t b3=(3/2)t+2d2=(3/2)t+2t=(7/2)t
a3/b3=(13/2)t/[(7/2)t]=13/7
这个方法可以解任意an/bn
收起