计算：1−121+2−122+3−123+…+100−12100．

Question

计算：(1−1)/(21)+(2−1)/(22)+(3−1)/(23)+…+(100−1)/(2100)．

Answer 1

设(1−1)/(21)+(2−1)/(22)+(3−1)/(23)+…+(100−1)/(2100)=x ①，
则(1)/(2)x=(2−1)/(23)+(3−1)/(24)+(4−1)/(25)+…+(100−1)/(2101)②，
①-②，得
(1)/(2)x=(1)/(22)+(1)/(23)+(1)/(24)+…+(1)/(2100)+(99)/(2101)
根据等比数列的求和公式，可得
(1)/(2)x＝(1)/(2)−(1)/(2100)+(99)/(2101)，
∴x＝1−(1)/(299)+(99)/(2100)．