Question

14．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0.01．

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）当精确到0.1时，2.03⁶ =（2+0.03）⁶=${C}_{6}^{0}$•2⁶+${C}_{6}^{1}$•2⁵•0.03+${C}_{6}^{2}$•2⁴•0.03²+…+${C}_{6}^{6}$•0.03⁶ 
≈${C}_{6}^{0}$•2⁶+${C}_{6}^{1}$•2⁵•0.03+${C}_{6}^{2}$•2⁴•0.03²=64+192×0.03+240×0.0009=64+5.76+0.216≈71.0．
即 2.03⁶ ≈71.0．
（2）当精确到0.01，2.03⁶ =（2+0.03）⁶=${C}_{6}^{0}$•2⁶+${C}_{6}^{1}$•2⁵•0.03+${C}_{6}^{2}$•2⁴•0.03²+…+${C}_{6}^{6}$•0.03⁶
≈${C}_{6}^{0}$•2⁶+${C}_{6}^{1}$•2⁵•0.03+${C}_{6}^{2}$•2⁴•0.03²+${C}_{6}^{3}$•2³•0.03³=64+192×0.03+240×0.0009+0.00432
=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．