Question

等差数列问题 谢谢

Answer 1

如图

Answer 2

Sn=2an-2^(n+1)
S(n-1)=2a(n-1)-2^n
an=Sn-S(n-1)
an=2an-2a(n-1)-2^n
an=2a(n-1)+2^n
等式两边除以2^n得
an/2^n=a(n-1)/2^(n-1)+1
[an/2^n]-[a(n-1)/2^(n-1)]=1
bn=an/2^n 则 b(n-1)=a(n-1)/2^(n-1)
bn-b(n-1)=1
{bn}是等差数列
2)S1=a1=4
bn=b1+(n-1)=(a1/2)+(n-1)=2+(n-1)=n+1
bn=n+1
Cn=bn*2^(-n)=(n+1)*(1/2)^n
Tn=2*(1/2)+3*(1/2)^2+....+(n+1)*(1/2)^n
1/2Tn=2*(1/2)^2+3*(1/2)^3+....n*(1/2)^n+(n+1)*(1/2)^(n+1)
两式相减得
1/2Tn=1+(1/2)^2+...+(1/2)^n-(n+1)*(1/2)^(n+1)
1/2Tn=[1-(1/2)^n]/(1-1/2)-[(n+1)*(1/2)^(n+1)]
1/2Tn=2[1-(1/2)^n]-[(n+1)*(1/2)^(n+1)]
Tn=4[1-(1/2)^n]-[(n+1)*(1/2)^n]
Tn=4-[(n+5)*(1/2)^n]