由an+Sn=2n得
an+1+Sn+1=2(n+1)
两式相减得2an+1-an=2,即2an+1-4=an-2,即an+1-2=
1 |
2 |
是首项为a1-2=-1,公比为
1 |
2 |
1 |
2 |
1 |
2 |
(Ⅱ)由(Ⅰ)知bn=(2−n)•(−1)•(
1 |
2 |
1 |
2 |
由bn+1−bn=
n−1 |
2n |
n−2 |
2n−1 |
n−1−2n+4 |
2n |
3−n |
2n |
由bn+1-bn<0得n>3,所以b1<b2<b3=b4>b5>…>bn
故bn的最大项为b3=b4=
1 |
4 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
n−1 |
2n |
n−2 |
2n−1 |
n−1−2n+4 |
2n |
3−n |
2n |
1 |
4 |