题目内容
17.执行如图所示的程序框图,则输出的结果为( )A. | 1006 | B. | 1008 | C. | 2015 | D. | 2016 |
分析 根据程序框图进行模拟计算即可.
解答 解:当i是奇数时,i2i2的余数是1>0,S=12+…+i2,
当i是偶数时,i2i2的余数0>0不成立,S=-22-…-i2,
故程序是计算S=1201512015(12-22+32-42+…-20142+20152)
=1201512015[1+(32-22)+…+(20152-20142)]
=1201512015(1+5+9+…+4029)
=1201512015×(1+4029)×10082(1+4029)×10082
=1201512015×4030×100824030×10082=1008,
故选:B
点评 本题主要考查程序框图的识别和判断,根据条件读懂程序是解决本题的关键.
分组 | 151.5~158.5 | 158.5~165.5 | 165.5~172.5 | 172.5~179.5 |
频数 | 6 | 21 | 27 | 6 |
频率 | 0.1 | 0.35 | a | 0.1 |
分组 | [29.86,29.90) | [29.90,29.94) | [29.94,29.98) | [29.98,30.02) | [30.02,30.06) | [30.06,30.10) | [30.10,30.14) |
频数 | 29 | 71 | 85 | 159 | 76 | 62 | 18 |
分组 | [29.86,29.90) | [29.90,29.94) | [29.94,29.98) | [29.98,30.02) | [30.02,30.06) | [30.06,30.10) | [30.10,30.14) |
频数 | 12 | 63 | 86 | 182 | 92 | 61 | 4 |