Question

已知等比数列{an}的前n项和为Sn,且an是Sn与2的等差中项,等差数列{bn}中,b1=2,点P(bn,bn+1)在

Answer 1

问题1:2a1=S1+2=a1+2

→a1=2

2a2=S2+2=a1+a2+2

→a2=4

问题2:an是Sn与2的等差中项

→2an=Sn+2

2a(n-1)=S(n-1)+2

两式相减

2an-2a(n-1)=Sn-S(n-1)=an

→an=2a(n-1)

→an=2a(n-1)=2^2a(n-2)=……＝2^(n-1)a1=2^n

点P(bn,bn+1)在直线y=x+2上

→b(n+1)=bn+2

→bn=b(n-1)+2=b(n-2)+2*2=b(n-3)+2*3=……=b1+2(n-1)=2+2(n-1)=2n

问题3:设Cn=an*bn=n*2^(n+1)

Tn=1*2^2+2*2^3+3*2^4+……+n*2^(n+1)

→2Tn= 1*2^3+2*2^4+……+(n-1)*2^(n+1)+n*2^(n+2)

两式相减

-Tn=Tn-2Tn=1*2^2+1*2^3+1*2^4+……+1*2^(n+1)-n*2^(n+2)

=4+8+16+……+2^(n+1)-n*2^(n+2)

=4+4+8+16+……+2^(n+1)-n*2^(n+2)-4

=8+8+16+……+2^(n+1)-n*2^(n+2)-4

=……

=2^(n+2)-n*2^(n+2)-4

=-(n-1)*2^(n+2)-4

故Tn=(n-1)*2^(n+2)+4