作业帮求和1/1*5 + 1/3*7 +1/5*9 +1/7*11+.+1/(2n-1)(2n+3)
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/07 21:53:57
求和1/1*5 + 1/3*7 +1/5*9 +1/7*11+.+1/(2n-1)(2n+3)
如题 顺便说明一下原理
=1/3-(n+1)/(2n+1)(2n+3)
=n(4n+5)/3(2n+1)(2n+3) 请问这两步是怎么回事？
如题 顺便说明一下原理
=1/3-(n+1)/(2n+1)(2n+3)
=n(4n+5)/3(2n+1)(2n+3) 请问这两步是怎么回事？
1/(2n-1)(2n+3)=1/4*[1/(2n-1)-1/(2n+3)]
所以
1/1*5+1/3*7+1/5*9+1/7*11+.1/(2n-1)*(2n+3)
=1/4*(1-1/5)+1/4*(1/3-1/7)+1/4*(1/5-1/9)+1/4*(1/7-1/11)+1/4*[1/(2n-1)-1/(2n+3)]
=1/4*[(1-1/5)+(1/3-1/7)+(1/5-1/9)+(1/7-1/11)+.+1/(2n-1)-1/(2n+3)]
=1/4*[1+1/3+1/(2n+1)-1/(2n+3)]
=1/3-(n+1)/(2n+1)(2n+3)
=n(4n+5)/3(2n+1)(2n+3)
所以
1/1*5+1/3*7+1/5*9+1/7*11+.1/(2n-1)*(2n+3)
=1/4*(1-1/5)+1/4*(1/3-1/7)+1/4*(1/5-1/9)+1/4*(1/7-1/11)+1/4*[1/(2n-1)-1/(2n+3)]
=1/4*[(1-1/5)+(1/3-1/7)+(1/5-1/9)+(1/7-1/11)+.+1/(2n-1)-1/(2n+3)]
=1/4*[1+1/3+1/(2n+1)-1/(2n+3)]
=1/3-(n+1)/(2n+1)(2n+3)
=n(4n+5)/3(2n+1)(2n+3)
求和：1-3+5-7+9-11+（-1）^n-1*(2n-1)=多少
n(n+1)(n+2)数列求和
一道数列求和题1/2n+3/4n+5/8n+...+（2n-1）/n*2^n
求和：Sn=1-3x+5x^2-7x^3+.+(2n+1)(-x)^n(n属于N*)
数列求和3+5+7+...+2n-1
数列求和：1*3+3*5+5*7+.+(2n-1)(2n+1)
各路大神请指教 级数求和 1/n(n+1)(n+2)(n+3).(n+k)
数列求和：Sn＝-1+3-5+7-…+(（-1）^n)(2n-1)
sn=1×3+2×5+3×7+...+n×(2n+1)数列求和
数列求和：1/2+3/4+5/8+7/16+.+(2n-1)/(2^n)
求和：S=1/2+3/4+5/8+7/16+.+（2n-1)/2^n
求和：1/3+7/4+5/8+...+(2n-1)/2^n