Sn=1-1/2+1/3-1/4+……+1/(2n-1)-1/2n
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/14 04:06:36
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?)
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!
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!
求和sn=1×2×3+2×3×4+……+n(n+1)(n+2)
已知Sn=2+5n+8n^2+…+(3n-1)n^n-1(n∈N*)求Sn
求和:Sn=1*n+2*(n-1)+3*(n-2)+……+n*1
已知数列{an}的前n项和为Sn=1+2+3+4+…+n,求f(n)= Sn /(n+32)Sn+1的最大值
已知:Sn=1+1/2+1/3+……+1/n,用数学归纳法证明:Sn^2>1+n/2(n>=2,n∈N+)
求和:Sn=1*2*3+2*3*4+……+n(n+1)(n+2)
1+2+3+4+.+n,求Sn
Sn=1x2+3x2^2+5x2^3+…+(2n-1)x2^n sn=2sn-sn
Sn=n(n+2)(n+4)的分项等于1/6[n(n+2)(n+4)(n+5)-(n-1)n(n+2)(n+4)]吗?
数列求和:Sn=-1+3-5+7-…+((-1)^n)(2n-1)
已知等差数列{an}的前n项和为Sn,且(2n-1)Sn+1 -(2n+1)Sn=4n²-1(n∈N*)
设数列{an}的前n项和为sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*)