Question

Sn是正项数列{an}的前n项和,且Sn=(1/4)an^2+(1/2)an-3/4

an=2n+1.已知bn=2^n,求Tn=a1b1+a2b2+....+anbn的值。

Answer 1

Sn=(1/4)an^2+(1/2)an-3/4S(n-1)=(1/4)a(n-1)^2+(1/2)a(n-1)-3/4Sn-S(n-1)=an(1/4)an^2+(1/2)an-[(1/4)a(n-1)^2+(1/2)a(n-1)]=an又an=2n+1解得a(n-1)=2n-1或-2n-1(负数，舍去）又a1=S1，解得a1=3或-1（舍去）所以an是以3为首项的，公差为2的等差数列anbn=(2n+1)2^n=n*2^(n+1)+2^nTn=1*2^2+2*2^3+....+n*2^(n+1)+2+2^2+2^3+....+2^n令M=1*2^2+2*2^3+....+n*2^(n+1)则2M=1*2^3+2*2^4+...n*2^(n+2)2M-M=n*2^(n+2)-2^2-2^3-2^4-...-2^(n+1)Tn=n*2^(n+2)-2^(n+1)+2

Answer 2

anbn=(2n+1)2^n=n*2^(n+1)+2^nTn=1*2^2+2*2^3+....+n*2^(n+1)+2+2^2+2^3+....+2^n令M=1*2^2+2*2^3+....+n*2^(n+1)则2M=1*2^3+2*2^4+...n*2^(n+2)2M-M=n*2^(n+2)-2^2-2^3-2^4-...-2^(n+1)Tn=n*2^(n+2)-2^(n+1)+2