On sums of powers of natural numbers
Fuente:
arXiv
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| Autore principale: | |
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| Natura: | Preprint |
| Pubblicazione: |
2024
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| _version_ | 1866913580233785344 |
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| author | Samsonadze, Eteri |
| author_facet | Samsonadze, Eteri |
| contents | The problem of finding the sum of a polynomial's values is considered. In particular, for any $n\geq 3$, the explicit formula for the sum of the $n$th powers of natural numbers $S_n=\sum_{x=1}^{m}x^{n}$ is proved: $$\sum_{x=1}^{m}x^{n}=(-1)^{n}m(m+1)(-\frac{1}{2}+\sum_{i=2}^{n}a_i(m+2)(m+3)...(m+i)),$$ here $a_i=\frac{1}{i+1}\sum_{k=1}^{i}\frac{(-1)^{k}k^{n}}{k!(i-k)!}$, $(i=2,3,...,n-1)$, $a_n=\frac{(-1)^n}{n+1}$. Note that this formula does not contain Bernoulli numbers. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2411_11859 |
| institution | arXiv |
| publishDate | 2024 |
| record_format | arxiv |
| spellingShingle | On sums of powers of natural numbers Samsonadze, Eteri General Mathematics 40G99, 05A19, 11B68, 05A10 The problem of finding the sum of a polynomial's values is considered. In particular, for any $n\geq 3$, the explicit formula for the sum of the $n$th powers of natural numbers $S_n=\sum_{x=1}^{m}x^{n}$ is proved: $$\sum_{x=1}^{m}x^{n}=(-1)^{n}m(m+1)(-\frac{1}{2}+\sum_{i=2}^{n}a_i(m+2)(m+3)...(m+i)),$$ here $a_i=\frac{1}{i+1}\sum_{k=1}^{i}\frac{(-1)^{k}k^{n}}{k!(i-k)!}$, $(i=2,3,...,n-1)$, $a_n=\frac{(-1)^n}{n+1}$. Note that this formula does not contain Bernoulli numbers. |
| title | On sums of powers of natural numbers |
| topic | General Mathematics 40G99, 05A19, 11B68, 05A10 |
| url | https://arxiv.org/abs/2411.11859 |