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)+...1⼀2n

如题... 快
2024-12-14 19:42:09
推荐回答(1个)
回答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