1、
[a2+a(2n+2)]/[b3+b(2n+1)]
=[2a(n+2)]/[2b(n+2)]
=a(n+2)/b(n+2)
=S[2(n+2)-1]/T[2(n+2)-1]
=S(2n+3)/T(2n+3)
S(2n+3)/T(2n+3)=(3n-2)/(5n+6)
令n=3,得
S9/T9=(3×3-2)/(5×3+6)=7/21=1/3
2、
设公差为d
Sn=na1+n(n-1)d/2
=na1+(n²-n)d/2
=na1+dn²/2 -dn/2
=(d/2)n²+(1/2)(2a1-d)n,即Sn的表达式没有常数项.
又Sn=(2n-1)(n+p)=2n²+(2p-1)n-p
-p=0
p=0