Question

20．根据下列要求的精确度，求2.03⁶的近似值．
（1）精确到0.1；
（2）精确到0.001．

Answer 1

分析 根据2.03⁶ =（2+0.03）⁶，按照二项式定理展开，结合精度求得它的值．

解答 解：（1）精确到0.1时，2.03⁶=（2+0.03）⁶=${C}_{6}^{0}$×2⁶+${C}_{6}^{1}$×2⁵×0.03+…+${C}_{6}^{6}$×0.03⁶
≈${C}_{6}^{0}$•2⁶+${C}_{6}^{1}$•2⁵×0.03+${C}_{6}^{2}$×2⁴×0.03²=64+5.76+0.216≈71.0．
即 2.03⁶≈71.0．
（1）精确到0.01时，2.03⁶=（2+0.03）⁶=${C}_{6}^{0}$×2⁶+${C}_{6}^{1}$×2⁵×0.03+…+${C}_{6}^{6}$×0.03⁶
≈${C}_{6}^{0}$•2⁶+${C}_{6}^{1}$•2⁵×0.03+${C}_{6}^{2}$×2⁴×0.03²+${C}_{6}^{3}$×2³×0.0³=64+5.76+0.216+0.00432≈70.98．
即 2.03⁶≈70.98．

点评 本题主要考查二项式定理的应用，精确度的定义，属于基础题．