用数学归纳法证明:1-1⼀2+1⼀3-1⼀4+...+1⼀2n-1-1⼀2n=1⼀n+1+1⼀n+2+...+1⼀n+1

n=1时，左=1-1/2=1/2 右面=1/2成立，
假设n=k时，成立：1-1/2+1/3-1/4+...+1/2k-1-1/2k=1/k+1+1/k+2+...+1/k+k
则n=k+1时，
右=1/(k+2)+1/(k+3)+...+1/(k+1+k)+1/(2K+2)
=1/(k+2)+1/k+3)+...+1/(2k+1)+1/(2k+2).........................1
左=[1-1/2+1/3-1/4+...+1/2k-1-1/2k]+1/(2k+1)-1/(2k+2)
=1/(k+1)+1/(k+2)+...+1/(k+k)+1/(2k+1)-1/(2k+2)
=1/(k+2)+1/(k+3)+...+(2k+1)+1/(k+1)-1/(2k+2)
=1/(k+2)+1/(k+3)+...+(2k+1)+(2k+2-k-1)/[(k+1)(2k+2)]
=1/(k+2)+1/(k+3)+...+(2k+1)+(k+1)/[(k+1)(2k+2)]
=1/(k+2)+1/(k+3)+...+(2k+1)+1/(2k+2).............................2
1式=2式
所以n=k+1时也成立，
所以原式成立。

1：显然n=1成立，n＝2也成立
2：假设n=k成立，有1-1/2+1/3-1/4+...+1/(2k-1)-1/2k=1/(k+1)+1/(k+2)+...+1/2k
当n=k+1时,左边＝1-1/2+1/3-1/4+...+1/(2k-1)-1/2k+[1/(2k+1)-1/(2k+2)]
＝1/(k+1)+1/(k+2)+...+1/2k+[1/(2k+1)-1/(2k+2)] (首位两项做减法有）
＝1/(k+2)+...+1/（2k+1)+1/(2k+2)
所以n=k+1时也成立。证毕