问题
填空题
(理).已知an=
|
答案
∵an=
1 |
4n+2100 |
∴a100-n=
4n |
2100(2100+4n) |
∴an +a100-n =
1 |
2100 |
∴S99=a1+a2+…+a99①
S99=a99+a98+…+a1②
①+②
2S99=99×
1 |
2100 |
∴S99=
99 |
2101 |
故答案为:
99 |
2101 |
(理).已知an=
|
∵an=
1 |
4n+2100 |
∴a100-n=
4n |
2100(2100+4n) |
∴an +a100-n =
1 |
2100 |
∴S99=a1+a2+…+a99①
S99=a99+a98+…+a1②
①+②
2S99=99×
1 |
2100 |
∴S99=
99 |
2101 |
故答案为:
99 |
2101 |