小升初算术题,(1/1+2)+(1/1+2+3)+……+(1/1+2+3+4+……+100)=(4/1×3)+(8/3×5)+(12/5×7)+……+﹙400/19×21)=(1-1/2-1/3-1/4-1/5)×(1+1/2+1/3+1/4+1/5+1/6)-(1-1/2-1/3-1/4-1/5-1/6)×(
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![小升初算术题,(1/1+2)+(1/1+2+3)+……+(1/1+2+3+4+……+100)=(4/1×3)+(8/3×5)+(12/5×7)+……+﹙400/19×21)=(1-1/2-1/3-1/4-1/5)×(1+1/2+1/3+1/4+1/5+1/6)-(1-1/2-1/3-1/4-1/5-1/6)×(](/uploads/image/z/6899333-5-3.jpg?t=%E5%B0%8F%E5%8D%87%E5%88%9D%E7%AE%97%E6%9C%AF%E9%A2%98%2C%EF%BC%881%EF%BC%8F1%EF%BC%8B2%EF%BC%89%EF%BC%8B%EF%BC%881%EF%BC%8F1%EF%BC%8B2%EF%BC%8B3%EF%BC%89%2B%E2%80%A6%E2%80%A6%2B%EF%BC%881%EF%BC%8F1%2B2%2B3%2B4%2B%E2%80%A6%E2%80%A6%2B100%EF%BC%89%3D%EF%BC%884%EF%BC%8F1%C3%973%EF%BC%89%EF%BC%8B%EF%BC%888%EF%BC%8F3%C3%975%EF%BC%89%EF%BC%8B%EF%BC%8812%EF%BC%8F5%C3%977%EF%BC%89%EF%BC%8B%E2%80%A6%E2%80%A6%EF%BC%8B%EF%B9%99400%EF%BC%8F19%C3%9721%EF%BC%89%3D%281-1%2F2-1%2F3-1%2F4-1%2F5%29%C3%97%281%2B1%2F2%2B1%2F3%2B1%2F4%2B1%2F5%2B1%2F6%29-%EF%BC%881-1%2F2-1%2F3-1%2F4-1%2F5-1%2F6%EF%BC%89%C3%97%EF%BC%88)
小升初算术题,(1/1+2)+(1/1+2+3)+……+(1/1+2+3+4+……+100)=(4/1×3)+(8/3×5)+(12/5×7)+……+﹙400/19×21)=(1-1/2-1/3-1/4-1/5)×(1+1/2+1/3+1/4+1/5+1/6)-(1-1/2-1/3-1/4-1/5-1/6)×(
小升初算术题,
(1/1+2)+(1/1+2+3)+……+(1/1+2+3+4+……+100)=
(4/1×3)+(8/3×5)+(12/5×7)+……+﹙400/19×21)=
(1-1/2-1/3-1/4-1/5)×(1+1/2+1/3+1/4+1/5+1/6)-(1-1/2-1/3-1/4-1/5-1/6)×(1+1/2+1/3+1/4+1/5)=
小升初算术题,(1/1+2)+(1/1+2+3)+……+(1/1+2+3+4+……+100)=(4/1×3)+(8/3×5)+(12/5×7)+……+﹙400/19×21)=(1-1/2-1/3-1/4-1/5)×(1+1/2+1/3+1/4+1/5+1/6)-(1-1/2-1/3-1/4-1/5-1/6)×(
第一个式子:1/(1+2)=1/3,1/(1+2)+1/(1+2+3)=1/3+1/6=1/2=2/4,1/(1+2)+1/(1+2+3)+1/(1+2+3+4)=1/2+1/10=3/5,以此类推,前n项相加,则分子为n,分母为n+2,则最终结果为:99/101
第二个式子:(个人认为最后一项系数错了,应该是40,而不是400)
4/(1*3)=2*(1-1/3),8/(3*5)=4*(1/3-1/5),12/(5*7)=6*(1/5-1/7),…… ,40/(19*21)=20(1/19-1/21),故以上各式相加,可得2*(1+1/3+1/5+……+1/19)-20/21 (能力问题,目前只能解到这了)
第三个式子:原式=(1-1/2-1/3-1/4-1/5)×(1+1/2+1/3+1/4+1/5)+(1-1/2-1/3-1/4-1/5)*(1/6)-[(1-1/2-1/3-1/4-1/5)×(1+1/2+1/3+1/4+1/5)-(1/6) *(1+1/2+1/3+1/4+1/5)]=(1-1/2-1/3-1/4-1/5)*(1/6)+(1/6)*(1+1/2+1/3+1/4+1/5)=1/6 * 2= 1/3