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 | | From: | Ben Bean | | Subject: | Any familiarity with this mathematical oddity? | | Date: | Sat, 15 Jan 2005 16:27:01 -0500 |
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 | Wow. What charm. Given any four consecutive whole numbers, the difference
between the cube and the square of the highest number, minus 3 times the
difference between the cube and the square of the next lower number, plus 3
times the difference between the cube and the square of the 2nd lowest
number, minus the difference between the cube and the square of the lowest
number; always equals six.
Know any of those famed Indian math enthusiasts, there downunder? Heard of
this one before? It's a gem, to be sure!
-Ben Bean
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| | From: | Nyrk | | Subject: | Re: Any familiarity with this mathematical oddity? | | Date: | Mon, 17 Jan 2005 16:53:56 GMT |
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| Ben Bean wrote:
> Wow. What charm. Given any four consecutive whole numbers, the difference
> between the cube and the square of the highest number, minus 3 times the
> difference between the cube and the square of the next lower number, plus 3
> times the difference between the cube and the square of the 2nd lowest
> number, minus the difference between the cube and the square of the lowest
> number; always equals six.
>
May be it's charming but there's no oddity.
What you say is that
((x+3)^3-(x+3)^2)-3*((x+2)^3-(x+2)^2)+3*((x+1)^3-(x+1)^2)-(x^3-x^2)=6
and it seems just the particular case for n=3 of the more general fact that:
G(x,n)=n!
where:
G(x,n)=Sum_k(C(k,n)*[(x+k)^n-(x+k)^(n-1)])
and C(k,n) are the binomial coefficents C(k,n) := n!/(k!*(n-k)!) and k!=
k*(k-1)*(k-2)*...*2*1
For instance:
(x+4)^4-(x+4)^3-4*((x+3)^4-(x+3)^3)+6*((x+2)^4-(x+2)^3)-4*((x+1)^4-(x+1)^3)+(x^4-x^3)=4!
Can u demonstrate it? :-)
Best regards.
> Know any of those famed Indian math enthusiasts, there downunder? Heard of
> this one before? It's a gem, to be sure!
>
> -Ben Bean
>
>
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| | From: | Nyrk | | Subject: | Re: Any familiarity with this mathematical oddity? | | Date: | Mon, 17 Jan 2005 16:55:46 GMT |
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| Errata Coorige:
an alternate sign is missing so the correct definition is:
G(x,n)=Sum_k((-1)^k * C(k,n)*[(x+k)^n-(x+k)^(n-1)])
Byez.
Nyrk wrote:
> Ben Bean wrote:
>
>> Wow. What charm. Given any four consecutive whole numbers, the difference
>> between the cube and the square of the highest number, minus 3 times the
>> difference between the cube and the square of the next lower number,
>> plus 3
>> times the difference between the cube and the square of the 2nd lowest
>> number, minus the difference between the cube and the square of the
>> lowest
>> number; always equals six.
>>
>
> May be it's charming but there's no oddity.
> What you say is that
> ((x+3)^3-(x+3)^2)-3*((x+2)^3-(x+2)^2)+3*((x+1)^3-(x+1)^2)-(x^3-x^2)=6
> and it seems just the particular case for n=3 of the more general fact
> that:
> G(x,n)=n!
> where:
> G(x,n)=Sum_k(C(k,n)*[(x+k)^n-(x+k)^(n-1)])
> and C(k,n) are the binomial coefficents C(k,n) := n!/(k!*(n-k)!) and k!=
> k*(k-1)*(k-2)*...*2*1
>
> For instance:
> (x+4)^4-(x+4)^3-4*((x+3)^4-(x+3)^3)+6*((x+2)^4-(x+2)^3)-4*((x+1)^4-(x+1)^3)+(x^4-x^3)=4!
>
>
> Can u demonstrate it? :-)
>
> Best regards.
>
>
>> Know any of those famed Indian math enthusiasts, there downunder?
>> Heard of
>> this one before? It's a gem, to be sure!
>>
>> -Ben Bean
>>
>>
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