Question

已知数列{an}中的前n项和为Sn=-3n^2+6n,数列{bn}满足bn=(1/2)^n-1，数列满足Cn=1/6an*bn,求｛an｝

已知数列{an}中的前n项和为Sn=-3n^2+6n,数列{bn}满足bn=(1/2)^n-1，数列｛cn｝满足Cn=1/6an*bn,求｛an｝的通项公式，求｛cn｝的前几项和Tn

Answer 1

当n=1时,a1=S1=-3×1²+6×1=3,
当n≥2时,
an=Sn-S<n-1>
=(-3n²+6n)-[-3(n-1)²+6(n-1)]
=-3n²+6n+3n²-6n+9-6n+6
=9-6n,
an=9-6n满足a1=3,
则｛an｝的通项公式：an=9-6n。

cn=1/6×(1/2)^(n-1)×(9-6n)=(3-2n)/2^n。
Tn=1/2^1+(-1)/2^2+(-3)/2^3+...+(5-2n)/2^(n-1)+(3-2n)/2^n，
有2Tn=1+(-1)/2^1+(-3)/2^2+(-5)/2^3+...+(3-2n)/2^(n-1)。
两式相减【2Tn-Tn】可得：
Tn=1+1/2^1×(-1-1)+1/2^2×[-3-(-1)]+1/2^3×[-5-(-3)]+...+1/2^(n-1)×[(3-2n)-(5-2n)]-(3-2n)/2^n
=1-2×[1/2^1+1/2^2+1/2^3+,..+1/2^(n-1)]-(3-2n)/2^n
=1-2×1/2[1-(1/2)^(n-1)]/(1-1/2)-(3-2n)/2^n
=1-2×[1-(1/2)^(n-1)]-(3-2n)/2^n
=[(2n+1)/2^n]-1，
即Tn=[(2n+1)/2^n]-1。