# Is there a pattern for reducing exponentiation to sigma sums?

The other day I was trying to find a method for cubing numbers similar to one I found for squaring numbers. I found that to find the square of a positive integer n, just sum up the first n odd integers.

$\sum_{t=1}^n 2t-1 = n^2$

Similarly, I found a method for cubing numbers

$\sum_{t=1}^n 3t^2-3t+1 = n^3$

Inside that, I realized I could condense 3t^2 to my sum I found earlier for squaring numbers, and I'd have a nested sigma sum. What I noticed at this point was that all I was doing was writing out in long hand the reduction of multiplication (and exponentiation) to the sum of 1, n times, which makes sense because after all, multiplication is just repeated addition. Also, the number of nested sigma sums was related to the power I was raising the original number to, which is also intuitive because it's just another series of additions.

What I'm curious about is if there is a pattern to this "reduction to summation" that I did. If I wanted to reduce a^b to a summation with terms that are at most of degree (b-1), how is there a repeating pattern that I could follow/extrapolate from the given sums that I have so far?

You exploited the fact that $$\sum_{t=1}^n (t^3-(t-1)^3)=n^3.$$ This result is clear, when you add up there is wholesale cancellation (telescoping). Your term $3t^2-3t+1$ is $t^3-(t-1)^3$.

Exactly the same idea works for any positive integer $b$. Use the fact that
$$\sum_{t=1}^n (t^b-(t-1)^b)=n^b.$$ Expand $(t-1)^b$ using the Binomial Theorem to get the analogue of your results for general $b$. The polynomial $t^b-(t-1)^b$ has degree $b-1$, precisely what you wanted.

For example, with $b=4$ we end up with $\sum_{t=1}^n (4t^3-6t^2+4t-1)$. With $b=5$ we get $\sum_{t=1}^n (5t^4-10t^3+10t^2-5t+1)$.

The procedure can indeed be used to build up to a formula for the sum of the first $n$ $b$-th powers.

The problem of summing consecutive powers has a long history. You might be interested in the Wikipedia article on Faulhaber's Formula.

• This is sometimes called the "fundamental theorem of discrete calculus," thinking of $t^b - (t - 1)^b$ as the discrete derivative of $t^b$. – Qiaochu Yuan Jun 29 '12 at 16:44
• Thank you very much. It always amazes me how applicable the binomial theorem is, and it shows up in something like this as well. – QtizedQ Jun 29 '12 at 17:58