题目内容
已知| C | 4 n |
| C | 5 n |
| C | 6 n |
| C | 12 n |
分析:利用
,
,
成等差数列,可得2
=
+
,求出n,再计算
的值.
| C | 4 n |
| C | 5 n |
| C | 6 n |
| C | 5 n |
| C | 4 n |
| C | 6 n |
| C | 12 n |
解答:解:∵
,
,
成等差数列,
∴2
=
+
,
∴2
=
+
,
∴
=
+
,
∴n2-21n+98=0,
∵n≥12,
∴n=14,
∴
=
=
=
=71.
故答案为:91.
| C | 4 n |
| C | 5 n |
| C | 6 n |
∴2
| C | 5 n |
| C | 4 n |
| C | 6 n |
∴2
| n! |
| 5!(n-5)! |
| n! |
| 4!(n-4)! |
| n! |
| 6!(n-6)! |
∴
| 2 |
| 5(n-5) |
| 1 |
| (n-4)(n-5) |
| 1 |
| 30 |
∴n2-21n+98=0,
∵n≥12,
∴n=14,
∴
| C | 12 n |
| C | 12 14 |
| C | 2 14 |
| 14×13 |
| 2 |
故答案为:91.
点评:本题考查组合数的计算,考查等差数列的性质,求出n是关键.
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