题目内容
求和W=
+4
+7
+10
+…+(3n+1)
.
| C | 0 n |
| C | 1 n |
| C | 2 n |
| C | 3 n |
| C | n n |
分析:通过等差数列与组合数的性质,运用反序求和方法,直接求出表达式的和.
解答:解:∵an=3n+1为等差数列,∴a0+an=a1+an-1=…,
而
=
,(运用反序求和方法),
∵W=
+4
+7
+…+(3n-2)
+(3n+1)
①,
=(3n+1)
+(3n-2)
+(3n-5)
+…+4
+
∴W=(3n+1)
+(3n-2)
+(3n-5)
+…+4
+
②,
①+②得2W=(3n+2)(
+
+
+…+
)=(3n+2)×2n,
∴W=(3n+2)×2n-1.
而
| C | k n |
| C | n-k n |
∵W=
| C | 0 n |
| C | 1 n |
| C | 2 n |
| C | n-1 n |
| C | n n |
=(3n+1)
| C | n n |
| C | n-1 n |
| C | n-2 n |
| C | 1 n |
| C | 0 n |
∴W=(3n+1)
| C | 0 n |
| C | 1 n |
| C | n-2 n |
| C | 1 n |
| C | 0 n |
①+②得2W=(3n+2)(
| C | 0 n |
| C | 1 n |
| C | 2 n |
| C | n n |
∴W=(3n+2)×2n-1.
点评:本题考查等差数列与组合数的性质,反序求和方法的应用,考查计算能力.
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