## Writing n'th powers as sums over lower powers

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### Writing n'th powers as sums over lower powers

This morning I came across this curious fact that the sum of the first n odd numbers equals to n². In other words:

sum_i=1..n (2i-1) = n²

Out of curiosity, I tried to find an analogous decomposition for n³. After some poking around, I found this:

sum_i=1..n (3i²-3i+1) = n³

I noticed that the coefficients of the summed term seem suspiciously familiar:

2, 1
3,-3, 1

This looks like the 2nd and 3rd rows of Pascal's triangle, except without the initial entry, and in alternating signs. To test my hypothesis, I asked WolframAlpha to solve the following sum:

sum_i=1..n (4i³-6i²+4i-1)

Guess what the answer was? That's right: n⁴.

The next sum continues this pattern:

sum_i=1..n (5i⁴-10i³+10i²-5i+1) = n⁵

Looks like my hypothesis was right: ne can be written as a sum of lower powers of n in alternating signs, the coefficients of which are taken from the n'th row of Pascal's Triangle except without the first entry. (The first row also trivially holds: sum_i=1..n (1) = n¹.)

Why this is so, however, eludes me. On the surface it looks suspiciously like solving for ne in the expansion of (n-1)e, however, that doesn't account for the summation sign. The formula without the summation, obviously, does not hold. So why is it that n'th powers can be decomposed exactly in this way? Anybody can explain why this is so?
quickfur
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### Re: Writing n'th powers as sums over lower powers

Turns out, the solution is very simple.

The whole idea is to formulate the difference between the p'th powers of adjacent numbers. That is, consider the sequence:

0^p, 1^p, 2^p, 3^p, 4^p, ... n^p

Call the difference between consecutive terms in this sequence D_i. That is:

D_1 = 1^p - 0^p = 1
D_2 = 2^p - 1^p
D_3 = 3^p - 2^p
...
etc.

In other words, we define D_i to be: i^p - (i-1)^p.

So if you start with (n-1)^p and add D_n to it, you'll get n^p. Because, starting from zero, adding D_1 gives you 1^p, then adding D_2 to that gives you 2^p, adding D_3 to that gives you 3^p, and so on, until you reach n^p. That is, by construction, sum_{i=1}^n D_i = n^p.

Now let's look at D_i more carefully:

D_i = i^p - (i-1)^p

We can expand (i-1)^p using binomial expansion to get:

(i-1)^p = c_0*i^p - c_1*i^(p-1) + c_2*i^(p-2) - c_3*i^(p-3) ... c_p*i^0

where the c_j's are just the binomial coefficients from the p'th row of Pascal's triangle. Substituting this back to the definition of D_i, we see that the i^p term gets cancelled out, leaving the 2nd and onwards coefficients of the lower power terms left, with alternating signs. So we get exactly the truncated binomial expansion in the sum as described in the OP. And by construction, when we sum up these D_i's from 1 to n, we obtain n^p.

QED.
quickfur
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