combinatorial argument

[jay_shark]

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 increasing exponent .

[T50_Omaha8]

I believe you can do this by the repitoire summing method for the next largest exponent, but I don't know if that's technically combinatorial (even though I learned it in combinatorics class).

[jay_shark]

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 .

[jay_shark]

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

[borisp]

I posted a general recursive method to find this type of sum, basically employing "summation by parts." Is this what you are looking for? It isn't slick, but it gets the job done.

http://forumserver.twoplustwo.com/showfl...ue#Post10807391

[jay_shark]

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 the right side . I think it's pretty neat because it's the only explicit way you can arrive at such a formula for these types of series .

[jay_shark]

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 second equation reduces to [(n+1)c2]^2 . Pretty interesting , nonetheless .