Question

（1）已知n∈N^*,求證：1+2+2²+2³+…+2^5n-1能被31整除；

（2）求0.998⁶的近似值，使誤差小于0.001.

Answer 1

(1) 證明略(2) 0.998⁶≈1-0.012=0.988

解析:

（1）證明  ∵1+2+2²+2³+…+2^5n-1

==2⁵ⁿ-1=32ⁿ-1                                                                                                                                3分

=（31+1）ⁿ-1

=31ⁿ+C·31^n-1+C·31^n-2+…+C·31+1-1

=31（31^n-1+C·31^n-2+…+C）                                                                                                               6分

顯然括號內(nèi)的數(shù)為正整數(shù)，

故原式能被31整除.                                                                                                                               7分

（2）解  ∵0.998⁶=（1-0.002）⁶

=1-C（0.002）+C（0.002）²-C（0.002）³+…                                                                                10分

第三項T₃=15×(0.002)²=0.000 06＜0.001,以后各項更小,∴0.998⁶≈1-0.012=0.988.                 14分