等差数列｛an｝｛bn｝的前项之和分别为Sn,Tn,Sn/Tn=2n-3/n+2,则a5/b4=.

Question

等差数列｛an｝ ｛bn｝的前项之和分别为 Sn,Tn,Sn/Tn=2n-3/n+2,则a5/b4=___________.

Answer 1

Sn=a1n+n(n-1)d1=(a1-d1)n+d1n^2
Tn=b1n+n(n-1)d2=(b1-d2)n+d2n^2
又有Sn/Tn=(2n-3)/(n+2)
所以(n+2)[(a1-d1)n+d1n^2]=(2n-3)[(b1-d2)n+d2n^2]
展开后,n的等次项相等,即
d1=2d2
2d1+2(a1-d1)=-3d2+2(b1-d2),即2a1=2b1-5d2
2(a1-d1)=-3(b1-d2)
都齐次,所以不妨设d1=2,则d2=1,a1=-1,b1=3
则a5=7,b4=6
a5/b4=7/6
其实方法挺基本的,算起来也不难,如果找不到巧方法,只要动手就好了.