作业帮裂项法怎么用1/(1*3)+1/(3*5)+1/(5*7)+…+1/[(2n-1)(2n+1)]以及(1/2)^0*(1/2)^1+(1/2)^1*(1/2)^2+…+(1/2)^(n-1)*(1/2)^n这两个式子求和,是要用裂项法吧,可是怎么弄来着.类似这种的裂项怎么玩啊~
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/24 06:30:16
![裂项法怎么用1/(1*3)+1/(3*5)+1/(5*7)+…+1/[(2n-1)(2n+1)]以及(1/2)^0*(1/2)^1+(1/2)^1*(1/2)^2+…+(1/2)^(n-1)*(1/2)^n这两个式子求和,是要用裂项法吧,可是怎么弄来着.类似这种的裂项怎么玩啊~](/uploads/image/z/10840635-27-5.jpg?t=%E8%A3%82%E9%A1%B9%E6%B3%95%E6%80%8E%E4%B9%88%E7%94%A81%2F%281%2A3%29%2B1%2F%283%2A5%29%2B1%2F%285%2A7%29%2B%E2%80%A6%2B1%2F%5B%282n-1%29%282n%2B1%29%5D%E4%BB%A5%E5%8F%8A%281%2F2%29%5E0%2A%281%2F2%29%5E1%2B%281%2F2%29%5E1%2A%281%2F2%29%5E2%2B%E2%80%A6%2B%281%2F2%29%5E%28n-1%29%2A%281%2F2%29%5En%E8%BF%99%E4%B8%A4%E4%B8%AA%E5%BC%8F%E5%AD%90%E6%B1%82%E5%92%8C%2C%E6%98%AF%E8%A6%81%E7%94%A8%E8%A3%82%E9%A1%B9%E6%B3%95%E5%90%A7%2C%E5%8F%AF%E6%98%AF%E6%80%8E%E4%B9%88%E5%BC%84%E6%9D%A5%E7%9D%80.%E7%B1%BB%E4%BC%BC%E8%BF%99%E7%A7%8D%E7%9A%84%E8%A3%82%E9%A1%B9%E6%80%8E%E4%B9%88%E7%8E%A9%E5%95%8A%7E)
裂项法怎么用1/(1*3)+1/(3*5)+1/(5*7)+…+1/[(2n-1)(2n+1)]以及(1/2)^0*(1/2)^1+(1/2)^1*(1/2)^2+…+(1/2)^(n-1)*(1/2)^n这两个式子求和,是要用裂项法吧,可是怎么弄来着.类似这种的裂项怎么玩啊~
裂项法怎么用
1/(1*3)+1/(3*5)+1/(5*7)+…+1/[(2n-1)(2n+1)]
以及
(1/2)^0*(1/2)^1+(1/2)^1*(1/2)^2+…+(1/2)^(n-1)*(1/2)^n
这两个式子求和,是要用裂项法吧,可是怎么弄来着.
类似这种的裂项怎么玩啊~
裂项法怎么用1/(1*3)+1/(3*5)+1/(5*7)+…+1/[(2n-1)(2n+1)]以及(1/2)^0*(1/2)^1+(1/2)^1*(1/2)^2+…+(1/2)^(n-1)*(1/2)^n这两个式子求和,是要用裂项法吧,可是怎么弄来着.类似这种的裂项怎么玩啊~
第一个是列项相消,利用公式:1/[(2n-1)(2n+1)=(1/2)*{1/(2n-1)-1/(2n+1)}
所以原式=0.5*{1-1/(2n+1)}=n/(2n+1)
列项相消就是把原来式子里的每部分拆开,使前一项中的某一项能够和后一项中的某一项相抵消,从而得到最后的计算结果也就是没有被抵消的几项的相加减.
第2个是用等比求和可直接求出来,是以0.5为第一项.0.25为公比的等比和
类似的列项还有1/(n-1)n(n+1)..
1/(1*3)+1/(3*5)+1/(5*7)+…+1/[(2n-1)(2n+1)]
=1/2(1-1/3+1/3-1/5+....+1/(2n+1)
=n/(2n+1)
(1/2)^0*(1/2)^1+(1/2)^1*(1/2)^2+…+(1/2)^(n-1)*(1/2)^n
=1/2+(1/2)^3+(1/2)^5...+(1/2)^(2n-1)
=(2^n-1)/3*2^(n-1)