Sn=2an-2,S(n+1)=2a(n+1)-2,两式做差，得a(n+1)=2ann=1时，S1=a1=2a1-2,所以，a1=2,an=2^n将p点坐标代入，bn-b(n+1)+2=0,即b(n+1)-bn=2,n项叠加，bn-b1=2(n-1),所以，bn=2n-1,(n>1)综上，an=2^n,bn=2n-1tn=2(2-1)+2^2(4-1)+~~~+2^n(2n-1)
2tn=2^2(2-1)+~~~+2^n(2n-3)+2^(n+1)(2n-1)
(此处为错位相减）所以，tn=(2n-3)2^(n+1)+6^为乘方。... |
