扩缩估算法求近似值：1102+1112+1122+…+110002的误差小于0.006的近似值是．

Question

扩缩估算法求近似值：(1)/(102)+(1)/(112)+(1)/(122)+…+(1)/(10002)的误差小于0.006的近似值是______．

Answer 1

(1)/(102)+(1)/(112)+(1)/(122)+…+(1)/(10002)，
≈(1)/(9×11)+(1)/(10×12)+(1)/(11×13)+…+(1)/(999×1001)，
=(1)/(2)×（(1)/(9)-(1)/(11)+(1)/(10)-(1)/(12)+(1)/(11)-(1)/(13)+…+(1)/(999)-(1)/(1001)），
=(1)/(2)×（(1)/(9)+(1)/(10)-(1)/(1000)-(1)/(1001)），
=(1)/(2)×(1883891)/(9009000)，
=(1883891)/(18018000)．
故答案为：(1883891)/(18018000)．