Question

已知等比数列{An}的前n项和为Sn=a·2^n+b.且a1=3.

Answer 1

an=Sn-S(n-1)=a*2^(n-1)
d=2,a1=a=3
Sn=3*(1-2^n)/(1-2)=a*2^n-3
b=-3,a=3
an=3*2^(n-1)
Bn=n/3*(1/2)^(n-1)
3Bn=n*(0.5)^(n-1)
3Tn=3B1+3B2+...+3Bn
= 1+2/2+3/4+...+(n-1)*0.5^(n-2)+n*0.5^(n-1)
6Tn=2+2/1+3/2+4/4+...+ n*0.5^(n-2)
相减得
3Tn=2+1/1+1/2+1/4+...+1*0.5^(n-2)-n*0.5^(n-1)
=2+1*[0.5^(n-1)-1]/(0.5-1)-n*0.5^(n-1)
=4-(n+2)*0.5^(n-1)