it should be:
(1/n)^n * (1 + 2 + 2^2 + 2^3 +....(n/2)+1 terms) = {2^(1 + n/2) - 1}*(1/n)^n
when n even
(1/n)^n * (1 + 2 + 2^2 + 2^3 +....(n-1/2)+1 terms) = {2^(1 + (n-1)/2) -
1}*(1/n)^n when n is odd
-Piyush
On Mon, Sep 12, 2011 at 8:01 PM, sandeep nagamalli <[email protected]>wrote:
> I think it is 1 / (2N)
>
> (1/N) * (1/N)*(N/2) = 1/(2N)
>
>
> On Mon, Sep 12, 2011 at 6:33 PM, Akash Mukherjee <[email protected]>wrote:
>
>> this is a case, but isnt der a case of circular permutation 2??
>> what say abt dis
>> 0,1
>> 2,3
>> 4,5
>> 0,1
>> 2,3
>> 4,5
>>
>> it wrks i gues??
>>
>>
>> On Mon, Sep 12, 2011 at 12:56 PM, teja bala <[email protected]>wrote:
>>
>>>
>>>
>>> I think it is 1/N
>>>
>>> let N=6 that means rand(6)= takes values 0-5 i.e 6 values.
>>> rand(m)=6 values
>>> rand(n)=6 values
>>> total combinations 6*6=36 values set but among dem array will change only
>>> for
>>> (0,0)(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)=6values of N
>>>
>>> So 6/36=1/6
>>>
>>> If we generalize it 1/N
>>>
>>> correct me if i'm wrong?
>>>
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>>
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>
>
>
> --
> Thanks&Regards:
> -sandeep
>
>
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