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Forum: Science, Math, and Philosophy

08-13-2007, 03:00 PM

Replies: 6

combinatorial argument

Posted By jay_shark

Re: combinatorial argument

Perhaps it is easier to show the following :

1^2+2^2+3^2+...+n^2 = 2*(n+1)c3+(n+1)c2 for n>=2

1^3+2^3+3^3+...+n^3 = 6*(n+1)c4 + 6*(n+1)c3 + (n+1)c2 for n>=3

Notice that the rhs of the...

Forum: Science, Math, and Philosophy

08-13-2007, 09:37 AM

Replies: 6

combinatorial argument

Posted By jay_shark

Re: combinatorial argument

Thx for the reply Boris , but that wasn't what I was looking for .

There is an argument that allows you to count something in two different ways . One way is the left side and the other way is...

Forum: Science, Math, and Philosophy

08-13-2007, 01:01 AM

Replies: 6

combinatorial argument

Posted By jay_shark

Re: combinatorial argument

Show that 1^4+2^4+3^4+...+n^4 = 24*(n+1)c5 +36*(n+1)c4 + 8*(n+1)c3 + 6*(n+1)c3 + (n+1)c2 for n>=4

Forum: Science, Math, and Philosophy

08-12-2007, 10:06 PM

Replies: 6

combinatorial argument

Posted By jay_shark

Re: combinatorial argument

The identity is n*(n+1)*(2n+1)*(3n^2+3n-1)/30 .

There is no direct way of getting this answer . Just find some polynomial of 5th degree that works out to the above after factoring .

Forum: Science, Math, and Philosophy

08-12-2007, 06:31 PM

Replies: 6

combinatorial argument

Posted By jay_shark

combinatorial argument

Use a combinatorial argument to find the identity of 1^4+ 2^4 +3^4 + 4^4 +....+n^4 .

In fact you can use a similar argument for any exponent you wish ,it just takes longer to do for each...

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