题目内容
数列
,
,…
的前n项和为( )
1 |
1+2 |
1 |
1+2+3 |
1 |
1+2+…+n |
A.
| B.
| C.
| D.
|
由数列可知数列的通项公式an=
=
=
=2(
-
),
∴数列的前n项和S=2(
-
+
-
+…+
-
)=2(
-
)=
,
故选:C.
1 |
1+2+…+(n+1) |
1 | ||
|
2 |
(n+1)(n+2) |
1 |
n+1 |
1 |
n+2 |
∴数列的前n项和S=2(
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
n+1 |
1 |
n+2 |
1 |
2 |
1 |
n+2 |
n |
n+2 |
故选:C.
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