This wiki page page inspired me to do discover \(\sum^n_{i=1}i^a\) with \(a\) being a natural number; I will just start using \(\sum^n_{i=1}i^0=1^0+2^0+3^0+...+n^0=n\) so I can discover the next ones; Using the same method that the linked page had done with \(n^2\) and \(n^3\), I will discover the polynomial that is equal to \(\sum^n_{i=1}i^a\);

First, I will use binomial theorem:

\((n+1)^{a+1}=\sum^{a+1}_{i=0}\binom{a+1}{i}n^i\) The \(1^{a+1-i}\) will be always 1, so it will not change the result if not written

Therefore:

\((n+1)^{a+1}=n^{a+1}+\sum^a_{i=0}\binom{a+1}{i}n^i\) (I took the "\(n^{a+1}\)" out of the sum notation)

\((n+1)^{a+1}-n^{a+1}=\sum^a_{i=0}\binom{a+1}{i}n^i\)

As the page did, I will substitute n by 1,2,3....n

\(2^{a+1}-1^{a+1}=\sum^a_{i=0}\binom{a+1}{i}1^i\)

\(3^{a+1}-2^{a+1}=\sum^a_{i=0}\binom{a+1}{i}2^i\)

\(4^{a+1}-3^{a+1}=\sum^a_{i=0}\binom{a+1}{i}3^i\)

\(\vdots\)

\(n^{a+1}-(n-1)^{a+1}=\sum^a_{i=0}\binom{a+1}{i}(n-1)^i\)

\((n+1)^{a+1}-n^{a+1}=\sum^a_{i=0}\binom{a+1}{i}n^i\)

Then I will sum everything, notice that everything on the left side, except \((n+1)^{a+1}\) and \(-1^{a+1}\) cancels themselves, for the right side, every sum has now another sum inside it!

\((n+1)^{a+1}-1=\sum^a_{i=0}\left (\binom{a+1}{i}\sum^n_{j=1}j^i \right )\)

I will work the left side now, remember what \((n+1)^{a+1}\) is, but now, we will have to take the \(\binom{a+1}{0}n^0=1\), because it will cancel with \(-1\):

\(\sum^{a+1}_{i=1}\binom{a+1}{i}n^i=\sum^a_{i=0}\left (\binom{a+1}{i}\sum^n_{j=1}j^i \right )\)

Now, I will take the term that appears on the sum of the right when i=a out of the notation:

\(\sum^{a+1}_{i=1}\binom{a+1}{i}n^i=\binom{a+1}{a}\sum^n_{j=1}j^a+\sum^{a-1}_{i=0}\left (\binom{a+1}{i}\sum^n_{j=1}j^i \right )\)

Notice that this term is exactally what we are searching, so I will isolate it:

\(\binom{a+1}{a}\sum^n_{j=1}j^a=\sum^{a+1}_{i=1}\binom{a+1}{i}n^i-\sum^{a-1}_{i=0}\left (\binom{a+1}{i}\sum^n_{j=1}j^i \right )\)

Remember that \(\binom{a+1}{a}=\binom{a+1}{1}=a+1\):

\(\sum^n_{j=1}j^a=\frac{\sum^{a+1}_{i=1}\binom{a+1}{i}n^i-\sum^{a-1}_{i=0}\left (\binom{a+1}{i}\sum^n_{j=1}j^i \right )}{a+1}\)

Now, Remember that, when \(a=0\), we have a polynomial expression \(n\), then if \(a=2\) also "is"(As in: there is a polynomial expression that has the same values) a polynomial expression, and, by a strong induction, if a is natural, then the expression \(\sum^n_{i=1}i^a\) also "is".

I also made a program, in Python, that returned the expression of \(\sum^n_{i=1}i^a\), but to much more values other than 0, 1, 2, 3(Which the original page did), some of the results are:

\(\sum^n_{i=1}i^{1}=\frac{n^2 + n}{2}\)

\(\sum^n_{i=1}i^{2}=\frac{2n^3 + 3n^2 + n}{6}\)

\(\sum^n_{i=1}i^{3}=\frac{n^4 + 2n^3 + n^{2}}{4}\)

\(\sum^n_{i=1}i^{4}=\frac{6n^5 + 15n^4 + 10n^3 - n}{30}\)

\(\sum^n_{i=1}i^{5}=\frac{2n^6 + 6n^5 + 5n^4 - 1n^2}{12}\)

\(\sum^n_{i=1}i^{6}=\frac{6n^7 + 21n^6 + 21n^5 - 7n^3 + n}{42}\)

\(\sum^n_{i=1}i^{7}=\frac{3n^{8} + 12n^{7} + 14n^{6} - 7n^{4} + 2n^{2}}{24}\)

\(\sum^n_{i=1}i^{8}=\frac{10n^{9} + 45n^{8} + 60n^{7} - 42n^{5} + 20n^{3} - 3n}{90}\)

\(\sum^n_{i=1}i^{9}=\frac{2n^{10} + 10n^{9} + 15n^{8} - 14n^{6} + 10n^{4} - 3n^{2}}{20}\)

There are patterns there, but I will make another note to that, as this is getting huge

## Comments

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TopNewestThis is amazing, i never knew we can generalize it !! Before reading this note , i used to approximate the answer.

\[ S_{k} = 1^{k}+2^{k}+3^{k}+\cdots+n^{k} =\displaystyle\sum_{r=1}^{n} r^{k} \]

For a large value of \( n \) (probably greater than \(1000\) for negligible error) we can convert the sum in an integral

\[ \displaystyle\sum_{r=0}^{n}r^{k} = n^{k+1} \int_{0}^{1} x^{k} dx \\ S_{k} \approx \dfrac{n^{k+1}}{k+1} \]

For example take \( k =1 \) and \( n = 2000 \)

\[ S_{1} \approx \dfrac{2000^{2}}{2} = 2000000 \\ \text{exact sum = } 2001000 \]

an error of only \( 1000 \) . the bigger the value of \( n \) , more accurate is the approximation – Sambhrant Sachan · 8 months, 1 week ago

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The term with the highest degree of the equivalent of \(\sum^n_{i=1}i^a\) is \(\frac{n^{a+1}}{a+1}\)

We can prove it by induction, as in: it works for \(a = 1\), because \(\sum^n_{i=1} i^1=\frac{n^2}{1+1} + ...\)

If it works for \(a = k, k-1,k-2, ...\), then for \(a = k + 1\), we have:

\(\sum^n_{i=1}i^{k+1}=\frac{\sum^{k+2}_{i=1}\binom{k+2}{i}n^i-\left(\sum^{a}_{i=0}\binom{k+2}{i}\sum^n_{j=1}j^i\right )}{k+2}\)

The term with the highest degree is \(\frac{n^{(k+1)+1}}{(k+1)+1}\) due the first sum highest degree, no term of the other sum has a equal or higher degree, because the highest one is \(\sum^n_{i=1}i^k\), which gives \(\frac{n^{k+1}}{k+1}\)

And as the others sums have a lower exponent, then they will have a lower degree of term(this is a strong induction) – Matheus Jahnke · 8 months, 1 week ago

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There's a (somewhat) general result for this involving Bernoulli numbers. It's called Faulhaber's formula. – Prasun Biswas · 8 months, 1 week ago

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– Matheus Jahnke · 8 months, 1 week ago

That is a very good comment, thanks for linking it, even though it make everything I did useless...Log in to reply