Question

若An和Bn分别表示数列{an}和{bn}前n项的和,对任意正整数n,an=-（2n+3）/2,4Bn-12An=13n

Answer 1

(1) 4bn-12an=13n

通项公式bn=13n/4+3an=13n/4-3(2n+3)/2=(n-18)/4

(2) kn=2^(n+1)*an=-(2n+3)/2*2^(n+1)=-(2n+3)*2^n

Sn=-[5*2+7*2^2+9*2^3+.+(2n+3)*2^n

2Sn=-[5*2^2+7*2^3+9*2^4+...+(2n+3)*2^(n+1)]

Sn-2Sn=-[6+2*2+2*2^2+2*2^3+.+2*2^n-(2n+3)*2^(n+1)]

-Sn=-[6+2*2*(2^n-1)/(2-1)-(2n+3)*2^(n+1)]

Sn=6+2*2^(n+1)-4-(2n+3)*2^(n+1)

=2-(2n+1)*2^(n+1)