数列an=2^n,抽取1、4、7...3n-2项后构成数列bn,求前n项和是?

consider
f(x) = (2^x)/ x^2
f'(x) = ln2 .2^x /x^2 - [2.2^x/x^3]
= 2^x(xln2-2)/x^3 =0
x= 2/ln2
=2.88
a2 = 1
a3= 8/9
min an = a3 =8/9
ie k=3

Cn的前n项和为Sn
Sn=a1b1+a2b2+...+anbn
=2^1*2+2^2*2*2+..+2^n*2n
2Sn= 2^2*2*1+ +2^n*2(n-1)+2^(n+1)*2n
两式相减：-Sn=2^1*2+2^2*2+.+2^n*2-2^(n+1)*2n
=2(2+2^2+..+2^n)-2^(n+1)*2n
=4(2^n-1)-n*2^(n+2)
=(1-n)*2^(n+2)-4
因此有Sn=(n-1)*2^(n+2)+4

数列{an}(an=2^n)为2,4,8,16...各数为偶数,不能相加得到奇数,故无法证明.应证{an},an=2^(n-1),为1,2,4,8,16...我用的是归纳法.n=1时,数列为1,Sn=1,任意项的和为1.n=2时,数列为1,2,Sn=3,任意项的和为1=1,2=2,1+2=3.n=...