1+2+…+n |
n |
n(n+1) |
2n |
n+1 |
2 |
1 |
anan+1 |
4 |
(n+1)(n+2) |
1 |
n+1 |
1 |
n+2 |
∴数列{bn}的前n项和=4[(
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
n+1 |
1 |
n+2 |
=4(
1 |
2 |
1 |
n+2 |
2n |
n+2 |
故答案为
2n |
n+2 |
1+2+…+n |
n |
1 |
anan+1 |
1+2+…+n |
n |
n(n+1) |
2n |
n+1 |
2 |
1 |
anan+1 |
4 |
(n+1)(n+2) |
1 |
n+1 |
1 |
n+2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
n+1 |
1 |
n+2 |
1 |
2 |
1 |
n+2 |
2n |
n+2 |
2n |
n+2 |