Question

已知数列an的前n项和为SN,且满足SN=1-an...

Answer 1

(1)当n=1,a1=S1=1-a1,所以a1=1/2
当n>=2时，
Sn=1-an
S{n-1}=1-a{n-1}
两式相减得，an=a{n-1}-an
即 an/a{n-1}=1/2
又S2=a1+a2=1-a2，所以a2=1/4
an=(1/4)(1/2)^(n-2)=(1/2)^n
当n=1时，1/2=a1
所以an=(1/2)^n
(2)bn=n/an=nx2^n,b1=2
Tn=b1+b2+b3+…+b{n-1}+bn
=2+2x2^2+3x2^3+…+(n-1)x2^(n-1)+nx2^n①
2Tn=2^2+2x2^3+3x2^4+…+(n-1)x2^n+nx2^(n+1)②
②-①得，Tn=-2-2^2-2^3-…-2^n +nx2^(n+1)
=-{[2(1-2^n)]/(1-2)}+nx2^(n+1)
=2+(n-1)x2^(n+1)