Question

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

如题... 快

Answer 1

1, n=1时，左边=1-1/2=1/2.右边=1/2成立
2，设n=k时成立就是 1-1/2+1/3-1/4+...+1/(2k-1)-1/2k=1/(k+1)+...1/(2k)
当 n=k+1时，则1-1/2+1/3+...+1/(2k-1)-1/2k+1/(2k+1)-1/(2k+2)=1/(k+1)+...1/(2k)+1/(2k+1)-1/(2k+2)=1/(k+2)+...+1/(2k)+1/(2k+1)+1/(k+1)-1/(2k+2)
下面证明 1/(k+1)-1/(2k+2)=1/(2k+2)???
1/(k+1)-1/(2k+2)=(2-1)/(2k+2)=1/(2k+2) ！！！
所以 1-1/2+1/3+...+1/(2k-1)-1/2k+1/(2k+1)-1/(2k+2) = 1/(k+1)+...1/(2k)+ 1/(2k+1)+1/（2k+2）就是说 n=k+1时成立所以对于一切n都会成立