题目内容
已知f(n)=
+
+…+
(n∈N+),则f(k+1)-f(k)=
+
+
-
+
+
-
.
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 3n-1 |
| 1 |
| 3k |
| 1 |
| 3k+1 |
| 1 |
| 3k+2 |
| 1 |
| k+1 |
| 1 |
| 3k |
| 1 |
| 3k+1 |
| 1 |
| 3k+2 |
| 1 |
| k+1 |
分析:由k到k+1时增加和减少的项即可求出.
解答:解:∵f(k)=
+
+…+
,
f(k+1)=
+
+…+
+
+
+
,
∴f(k+1)-f(k)=
+
+
-
.
故答案为
+
+
-
.
| 1 |
| k+1 |
| 1 |
| k+2 |
| 1 |
| 3k-1 |
f(k+1)=
| 1 |
| (k+1)+1 |
| 1 |
| (k+1)+2 |
| 1 |
| 3(k+1)-4 |
| 1 |
| 3(k+1)-3 |
| 1 |
| 3(k+1)-2 |
| 1 |
| 3(k+1)-1 |
∴f(k+1)-f(k)=
| 1 |
| 3k |
| 1 |
| 3k+1 |
| 1 |
| 3k+2 |
| 1 |
| k+1 |
故答案为
| 1 |
| 3k |
| 1 |
| 3k+1 |
| 1 |
| 3k+2 |
| 1 |
| k+1 |
点评:正确弄清由k到k+1时增加和减少的项是解题的关键.
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