當(dāng)n=1時(shí),a1=S1=-3×1²+6×1=3,
當(dāng)n≥2時(shí),
an=Sn-S
=(-3n²+6n)-[-3(n-1)²+6(n-1)]
=-3n²+6n+3n²-6n+9-6n+6
=9-6n,
an=9-6n滿足a1=3,
則｛an｝的通項(xiàng)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).
兩式相減【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.