Question

1/1*3+1/3*5+1/5*7+....+1/（2n-1）(2n+1)=n/(2n+1)

用数学归纳法证明

Answer 1

1/1*3=0.5*（1-1/3）=1/(1*2+1)
1/1*3+1/3*5=0.5*(1-1/3+1/3-1/5)=2/(2*2+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）-1/(2n+1)}
=1/2{1-1/(2n+1)}
=n/(2n+1)
1/1*3+1/3*5+1/5*7+....+1/（2n-1）(2n+1)+1/（2n+1）(2n+3)
=1/2{1-1/3+1/3-1/5+....+1/（2n-1）-1/(2n+1)+1/（2n+1）-1/(2n+3)}
=1/2{1-1/(2n+3)}
=(n+1)/[2(n+1) +1]
所以
1/1*3+1/3*5+1/5*7+....+1/（2n-1）(2n+1)=0.5*（1-1/3+1/3-1/5....-1/（2n+1）)
=n/(2n+1)

Answer 2

1/1*3=0.5*（1-1/3）
1/3*5=0.5*(1/3-1/5) ...
1/（2n-1）(2n+1)=0.5*(1/（2n-1）-1/（2n+1）)

所以
1/1*3+1/3*5+1/5*7+....+1/（2n-1）(2n+1)=0.5*（1-1/3+1/3-1/5....-1/（2n+1）)
=0.5*（1-1/（2n+1）)
=n/(2n+1)