题目内容
1×
+2×
+3×
+…+n×
=
-
•(
)n-1 -
•(
)n
-
•(
)n-1 -
•(
)n．
| 1 |
| 3 |
| 1 |
| 9 |
| 1 |
| 27 |
| 1 |
| 3n |
| 3 |
| 4 |
| 1 |
| 4 |
| 1 |
| 3 |
| n |
| 2 |
| 1 |
| 3 |
| 3 |
| 4 |
| 1 |
| 4 |
| 1 |
| 3 |
| n |
| 2 |
| 1 |
| 3 |
分析：通项是n×
，可看作等差数列{n}与等比数列{
}对应项相乘得到，可用错位相消法求和．
| 1 |
| 3n |
| 1 |
| 3n |
解答：解：Sn=1×
+2×
+3×
+…+n×
两边同乘以
得
Sn=1×
+2×
+ …+(n-1)
+n×
两式相减得
Sn=
+
+
+…+
-n×
=
-n×
=
(1-(
)n) -n×
∴Sn=
-
• (
)n-1 -
•(
)n
故答案为
-
• (
)n-1 -
•(
)n
| 1 |
| 3 |
| 1 |
| 9 |
| 1 |
| 27 |
| 1 |
| 3n |
两边同乘以
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 9 |
| 1 |
| 27 |
| 1 |
| 3n |
| 1 |
| 3n+1 |
两式相减得
| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 9 |
| 1 |
| 27 |
| 1 |
| 3n |
| 1 |
| 3n+1 |
=
| ||||
1-
|
| 1 |
| 3n+1 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3n+1 |
∴Sn=
| 3 |
| 4 |
| 1 |
| 4 |
| 1 |
| 3 |
| n |
| 2 |
| 1 |
| 3 |
故答案为
| 3 |
| 4 |
| 1 |
| 4 |
| 1 |
| 3 |
| n |
| 2 |
| 1 |
| 3 |
点评：本题考查数列求和的错位相消法．适用于形如{an•bn}前n项和，其中{an}是等差数列，{bn}是等比数列．解题格式比较固定，要注意计算的准确性．
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