题目内容
(2+
)+(4+
)+…+(2n+
)=
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2n |
n2+n+1-
| 1 |
| 2n |
n2+n+1-
.| 1 |
| 2n |
分析:仔细观察:(2+
)+(4+
)+…+(2n+
),根据其结构性质把它等价转化为2(1+2+…+n)+(
+
+…+
),然后利用等差数列和等比数列的前n项和公式进行求解.
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2n |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2n |
解答:解:(2+
)+(4+
)+…+(2n+
)
=2(1+2+…+n)+(
+
+…+
)
=2×
+
=n2+n+1-
.
故答案为:n2+n+1-
.
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2n |
=2(1+2+…+n)+(
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2n |
=2×
| n(n+1) |
| 2 |
| ||||
1-
|
=n2+n+1-
| 1 |
| 2n |
故答案为:n2+n+1-
| 1 |
| 2n |
点评:本题考查数列的前n项和公式的求法,解题时要认真审题,注意合理地进行等价转化,注意等差数列和等比数列前n项和公式的灵活运用.
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