一道数列题~~~~~~~~~~

1),
a(n+1)=2an+1,
知道a(n+1)+1=2(an+1)
所以a(n+1)+1/an+1=2,
bn=an+1
a1=1,b1=2,
bn是等比数列,等比系数是2，bn=2^n;
2),
bn=2^n;,
根据bn=an+1
an=2^n-1,

3),
cn=(n+1)/an+1=(n+1)/2^n
前n项和为
Sn=c1+c2+c3+.....+cn
Sn=(1+1)/2^1+(2+1)/2^2+(3+1)/2^3+....+(n+1)/2^n ①
两边同乘以1/2
1/2*Sn=(1+1)/2^2+(2+1)/2^3+(3+1)/2^4+.....+(n+1)/2^(n+1) ②
①-②得
Sn-1/2Sn=[(1+1)/2^1+(2+1)/2^2-(1+1)/2^2+(3+1)/2^3-(2+1)/2^3+(4+1)/2^4-(3+1)/2^4+....-n/2^n+(n+1)/2^n-(n+1)/2^(n+1)]
=2/2+1/2^2+1/2^3+1/2^4+.....+1/2^n-(n+1)/2^(n+1)
=1/2+1/2*(1-1/2^n)/(1-1/2)-(n+1)/2^(n+1)
=1/2+1-1/2^n-(n+1)/2^(n+1)
=3/2-1/2^n-(n+1)/2^(n+1)
所以
Sn=3-1/2^(n-1)-(n+1)/2^n

1、
bn=an+1＝2a（n-1)+1+1
= 2[ a（n-1)+1 ]
=2b(n-1)
b1=a1+1=2
所以｛bn｝是以2为首项，2为公比的等比数列

2、
｛bn｝是以2为首项，2为公比的等比数列
所以bn通项：bn=2*2^(n-1)
=2^n
故：an＝bn-1
＝2^n-1

3、cn=n+1/2^