A big sum

damoclark

Can anyone help me with this sum

1^10 + 2^10 + 3^10 ......... +998^10 + 999^10 + 1000^10 = ?

I read that when Gauss was a kid at school he solved the simplier problem of summing all the numbers in his head between 1 and 100, before the teacher and all the other kids, by the observing the fact that 100+1=101, 99+2= 101, 98 +3 =101, etc and that there are 50 such sums, which gives an answer of 50*101=5050 for the problem.

The above sum, with powers of 10, I heard that one of the Bernoulli brothers could solve within 10 minutes with just pen and paper.

Any ideas on how he did it? I've been trying to work out a fast way to do this sum, with integrals but am always getting stuck.

Gokul43201

It shouldn't be terribly hard (cumbersome, yes) to derive a formula for the sum of n-th powers. I think the easiest way to do this might be by employing a telescoping sum. I recall a method involving Stirling Numbers and permutations...perhaps you can even Google it ?

Hurkyl

It's known that the sum of the first n m-th powers will be a degree (m+1) polynomial in n. (In fact, a few of the terms are known)

I don't know how fast you could solve a 10x10 system of equations, though, to determine the coefficients. I know there are easier ways of deriving this, but I don't remember them.

shmoe

Try a search for the "Euler-Maclaurin summation formula", it's possible it could handle that sum by hand in 15 minutes, but I haven't tried. It involves the Bernoulli numbers and functions, so perhaps it was known to (one of) them. I confess ignorance on the history of the method though.

robert Ihnot

Jacques Bernoulli, around 1690, was evidently the first person to publish on this subject and says:

"With the help of [these formulas] it took me less than half of
a quarter of an hour to find that the 10th powers of the first
1000 numbers being added together will yield the sum

91409924241424243424241924242500"
(Hey damoclark! Is that the exact sum you are looking for?)

The reference is: http://www.mathpages.com/home/kmath279.htm.

If I remember right this was discussed in Newman's World of Mathematics
.

It is also known that the early terms, providing it goes that far, for nth power sums from 1 to X are: [tex]\frac{X^_(n+1)}{n+1}+\frac{X^n}{2}[/tex][tex]+\frac{nX^_(n-1)}{12}[/tex], etc., and all the terms can be constructed using Bernoulli numbers,see: http://numbers.computation.free.fr/C...bernoulli.html

damoclark

Thanks Shmoe, I'll check out the "Euler-Maclaurin summation formula".

So the Sum could be calculated with a polynomial of order 11, which would mean I would have to substitute the value of 10^3, for the polynomial variable, to calculate the answer.

I've been looking at the function [tex]y = x^{10}[/tex], and intergrating under that to get an approximation of the answer, ie [tex]\frac{x^{11}}{11}[/tex], so an over estimation of the sum would be,[tex]\frac{x^{11}}{11}[/tex], setting x=10^3, so I guess the answer has under 33 decimal digits.

The problem, now I,m having is finding a simple way to calculate the other co-efficients, without a calculator.

damoclark

Thanks for the links Robert.
So if you know the Bernoulli numbers or Bernoulli polynomial in advance you can write down the formula quickly, I assume.But I wonder if there's another way to do, maybe involving a infinite series thats converges fast....