Question

1×3分之1+3分之1×5分之1+5分之1×7分之1……+（2n-1）分之1×（2n+1）分之1

求n的大小，要详细过程

Answer 1

解：因为 1x3分之1 =(1-1/3)*1/2
3×5分之1=(1/3-1/5)*1/2
5×7分之1=(1/5-1/7)*1/2
。。。。。
（2n-1）分之1×（2n+1）分之1 =[1/(2n-1)-1/(2n+1)]*1/2
所以，原式=(1-1/3)*1/2 + (1/3-1/5)*1/2 + (1/5-1/7)*1/2 +.......+ [1/(2n-1)-1/(2n+1)]*1/2
=[(1-1/3) + (1/3-1/5) + (1/5-1/7) +.......+ 1/(2n-1)-1/(2n+1)]*1/2
=[1-1/3 + 1/3-1/5 + 1/5-1/7+.......+ 1/(2n-1)-1/(2n+1)]*1/2
=[1-1/(2n+1)]*1/2
=[2n/(2n+1)]*1/2
=n/(2n+1)

Answer 2

1x3分之1+3×5分之1+5×7分之1+(2n-1)x( 2n+1)分之1
=1/2X【（1-1/3）+（1/3-1/5）+（1/5-1/7）+..+（1/2n-1-1/2n+1)】
=1/2x（1-1/2n+1)
=1/2x(2n+1-1)/(2n+1)
=1/2x2n/(2n+1)
=n/2n+1