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- October 30th 2006, 06:58 AM #1

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- October 30th 2006, 07:44 AM #2

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- October 30th 2006, 10:18 AM #3

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Hello, ling_c_0202!

I'll show you a*primitive*method for this problem.

Can anyone tell me how to get a formula for:

. . , where is a positive integer.

Crank out the first few sums and take consecutive differences.

Then take differences of the differences, etc., until we get a constant sequence.

We get a constant sequence at the**third**differences.

This tells us that the generating function of of the**third**degree (a cubic).

We have the general cubic function: .

. . and we must determine

We will use the first four sums from our list.

.

Now we must solve the system of equations, but it's quite simple . . .

Back-substitute and get: .

Hence: .

Therefore: .

- November 1st 2006, 08:10 AM #4

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- November 1st 2006, 10:17 AM #5

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Hello, ling_c_0202 and all!

Here's another approach to Sums-of-powers formulas.

(I'll give you the 'long version'.)

If we know the "previous" sums, say: . . and .

. .we can find the "next" formula: .

Consider the next-higher power (3) and the difference of two consecutive cubes:

. .

Now let and "stack" the equations.

. . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Add the equations: . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

And we have: . . . . .

Solve for .

Factor: .

Therefore: .

- November 15th 2006, 10:14 AM #6

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- November 15th 2006, 12:48 PM #7

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- November 15th 2006, 07:38 PM #8

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- November 16th 2006, 06:44 AM #9

- November 16th 2006, 08:44 AM #10

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