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)+
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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?)
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!
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