题目内容
(1+
)+(1+
×2)+(1+
×3)+…(1+
×10)+(1+
×11)的结果是x,那么与x最接近的整数是
| 19 |
| 92 |
| 19 |
| 92 |
| 19 |
| 92 |
| 19 |
| 92 |
| 19 |
| 92 |
25
25
.分析:根据题目特点,原式变为1×11+(1+2+3+…+10+11)×
,括号内运用高斯求和公式求出结果,原式变为11+66×
,进一步计算即可.
| 19 |
| 92 |
| 19 |
| 92 |
解答:解:x=(1+
)+(1+
×2)+(1+
×3)+…+(1+
×10)+(1+
×11),
=1×11+(1+2+3+…+10+11)×
,
=11+(1+11)×11÷2×
,
=11+66×
,
=11+13.63,
=24.63;
因此,与x最接近的整数是25.
故答案为:25.
| 19 |
| 92 |
| 19 |
| 92 |
| 19 |
| 92 |
| 19 |
| 92 |
| 19 |
| 92 |
=1×11+(1+2+3+…+10+11)×
| 19 |
| 92 |
=11+(1+11)×11÷2×
| 19 |
| 92 |
=11+66×
| 19 |
| 92 |
=11+13.63,
=24.63;
因此,与x最接近的整数是25.
故答案为:25.
点评:此题考查了学生通过认真审题,运用运算技巧,灵活解答问题的能力.
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