Simone已知数列an的首项
的有关信息介绍如下：
1.a(n+1)=2sn+2^(n+1)-1
an=2s(n-1)+2^n-1
a(n+1)-an=2(sn-s(n-1))+2^(n+1)-2^n
a(n+1)=3an+2^(n+1)-2^n
a(n+1)+2^(n+1)=3an+2*2^(n+1)-2^n
a(n+1)+2^(n+1)=3(an+2^n)
b(n+1)=3bn,b1=1+2=3,
数列bn是首相为3，公比为3的等比数列
2.bn=3^n,an=3^n-2^n
cn=2^n/(1+2^n)(1+2^(n+1))=(2*2^n-2^n+1-1)/(2^n+1)(2^(n+1)+1)
cn=(2^(n+1)+1)/(2^(n+1)+1)(2^n+1)-(2^n+1)/(2^(n+1)+1)(2^n+1)
cn=1/(2^n+1)-1/(2^(n+1)+1)
c1+c2+c3+...+cn=1/3-1/5+1/5-1/9+...+1/(2^n+1)-1/(2^(n+1)+1)
=1/3-1/(2^(n+1)+1)
