题目内容
1+(
| 1 |
| ||
| 2 |
1+(
| 2 |
| ||
| 2 |
1+(
| 3 |
| ||
| 2 |
…
(1)请用含有n(n是正整数)的等式表示上述变化规律
Sn=
| ||
| 2 |
Sn=
;
| ||
| 2 |
(2)推算出OA10的长
| 10 |
| 10 |
(3)S12+S22+S32+…+S102的值等于
| 55 |
| 4 |
| 55 |
| 4 |
分析:(1)分别求出S1、S2、S3…Sn,找出规律即可;
(2)根据勾股定理求出OA1,OA2,…OA10即可;
(3)首先求出S12+S22+S32+…+Sn2的通项公式,然后把n=10代入即可.
(2)根据勾股定理求出OA1,OA2,…OA10即可;
(3)首先求出S12+S22+S32+…+Sn2的通项公式,然后把n=10代入即可.
解答:解:(1)1+(
)2=2,S1=
;1+(
)2=3,S2=
;1+(
)2=4,S3=
…Sn=
,
(2)OA2=
=
,OA3=
,…OA10=
;
(3)S12=
,S22=
,S32=
,…Sn2=
,
S12+S22+S32+…+Sn2=
+
+…+
=
,
当n=10时,S12+S22+S32+…+S102=
.
故答案为Sn=
,
,
.
| 1 |
| ||
| 2 |
| 2 |
| ||
| 2 |
| 3 |
| ||
| 2 |
| ||
| 2 |
(2)OA2=
| OA12+A1A22 |
| 2 |
| 3 |
| 10 |
(3)S12=
| 1 |
| 4 |
| 2 |
| 4 |
| 3 |
| 4 |
| n |
| 4 |
S12+S22+S32+…+Sn2=
| 1 |
| 4 |
| 2 |
| 4 |
| n |
| 4 |
| n(n+1) |
| 8 |
当n=10时,S12+S22+S32+…+S102=
| 55 |
| 4 |
故答案为Sn=
| ||
| 2 |
| 10 |
| 55 |
| 4 |
点评:本题主要考查勾股定理的知识点,解答本题的关键是熟练运用勾股定理,此题难度不大.
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