知等比数列{an}前n项和为Sn=a*2^n+b,且a1=3.⑴求a、b值及{an}通项公式;⑵设bn=n/an,求数列{bn}的前n项和Tn.

Question

知等比数列{an}前n项和为Sn=a*2^n+b,且a1=3.
⑴求a、b值及{an}通项公式;
⑵设bn=n/an,求数列{bn}的前n项和Tn.

Answer 1

Sn=a2^n+b
a1=2a+b=3
a1+a2=4a+b,a2=2a
a1+a2+a3=8a+b.a3=4a
等比数列,a2/a1=a3/a2
a=3,b=-3
an=3*2^(n-1)
用错位相减法：
bn=(2/3)*n/2^n
Tn=(2/3)*[1/2+2/2^2+...+n/2^n]
Tn/2=(2/3)*[[1/2^2+2/2^3+...+(n-1)/2^n+n/2^(n+1)]
相减：Tn/2=（2/3）*[1/2+1/2^2+...+1/2^n-n/2^(n+1)]
Tn=(4/3)*[1-(2+n)/2^(n+1)]