Sn＝1-1/2+1/3-1/4+……+1/（2n-1）-1/2nTn＝1/（n+1）+1/（n+2）+1/（n+3）+……+1/2n用归纳法证明Sn＝Tn我求的是Sk+1＝-1/2+1/3-1/4+……+1/（2k-1）-1/2k+1/（2k+1）-1/（2k+2）那么Tk＝1/（k+1）+1/（k+2）+1/（k+3）+
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/20 05:27:57
Sn＝1-1/2+1/3-1/4+……+1/（2n-1）-1/2nTn＝1/（n+1）+1/（n+2）+1/（n+3）+……+1/2n用归纳法证明Sn＝Tn我求的是Sk+1＝-1/2+1/3-1/4+……+1/（2k-1）-1/2k+1/（2k+1）-1/（2k+2）那么Tk＝1/（k+1）+1/（k+2）+1/（k+3）+
Sn＝1-1/2+1/3-1/4+……+1/（2n-1）-1/2n
Tn＝1/（n+1）+1/（n+2）+1/（n+3）+……+1/2n
用归纳法证明Sn＝Tn
我求的是Sk+1＝-1/2+1/3-1/4+……+1/（2k-1）-1/2k+1/（2k+1）-1/（2k+2）
那么Tk＝1/（k+1）+1/（k+2）+1/（k+3）+……+1/2k－1+1/（2k+1）-1/（2k+2）
可是Tk没有办法证出相等...谁知道怎么证,
还有,如果写Tk+1的话,为什么会少一个（1/2k）项呢.（用Tk+1是别人算的,我忘了抄式子了,但是,前面的1/（k+1)肯定会删掉但为什么还少个1/2k呢?）
Sn＝1-1/2+1/3-1/4+……+1/（2n-1）-1/2nTn＝1/（n+1）+1/（n+2）+1/（n+3）+……+1/2n用归纳法证明Sn＝Tn我求的是Sk+1＝-1/2+1/3-1/4+……+1/（2k-1）-1/2k+1/（2k+1）-1/（2k+2）那么Tk＝1/（k+1）+1/（k+2）+1/（k+3）+
假设Sk=Tk,显然因为Tk=1/(k+1)+1/(k+2)+...+1/(k+k)
所以：Tk+1=1/[(k+1)+1]+1/[(k+1)+2]+...+1/[(k+1)+k-1]+1/[(k+1)+k]+1/[(k+1)+k+1]
={1/(k+2)+1/(k+3)+...+1/(k+k) }+ 1/(2k+1)+1/[2(k+1)]
= { [Tk] -1/(k+1) } +1/(2k+1)+1/[2(k+1)]
（由归纳假设）=Sk+ {1/(2k+1)+1/[2(k+1)]- 1/(k+1)}=Sk+{1/[2(k+1)-1]-1/[2(k+1)}=Sk+1
归纳证明成立!