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I know that $$\sum^n_{i=1}i^2=\frac{1}{6}n(n+1)(2n+1)$$ and $$\sum^n_{i=1}i^3=\left(\sum^n_{i=1}i\right)^2.$$ Here is the question: is there a formula for $$\sum^n_{i=1}i^4.$$

Michael Rozenberg
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abc...
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  • Yes, in fact continuing the trend of your other examples, it is a polynomial in $n$ of order 5. If you calculate the first few terms, you should be able to figure it out. However you should be careful about notation in your sums, and not use $n$ both as the upper limit and as the number counting up. – mlk Sep 18 '17 at 10:18
  • Does it mean $\sum i^n = $ polynomial of degree n+1? – john doe Sep 18 '17 at 10:20
  • I just edited. Thank you. – abc... Sep 18 '17 at 10:22
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    you may want to read https://en.wikipedia.org/wiki/Faulhaber%27s_formula?wprov=sfla1 – Huang Sep 18 '17 at 10:25
  • @Huang thanks for that. – abc... Sep 18 '17 at 10:31
  • https://math.stackexchange.com/questions/1667258/how-is-faulhabers-formula-derived, https://math.stackexchange.com/questions/2035188/methods-to-compute-sum-k-1nkp-without-faulhabers-formula – Hans Lundmark Sep 18 '17 at 11:04

3 Answers3

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We can get the formula by the following way.

$$(n+1)^5-1=\sum_{k=1}^n((k+1)^5-k^5)=\sum_{k=1}^n(5k^4+10k^3+10k^2+5k+1).$$ Thus, $$\sum_{k=1}^nk^4=$$ $$=\frac{1}{5}\left((n+1)^5-1-10\cdot\frac{n^2(n+1)^2}{4}-10\cdot\frac{n(n+1)(2n+1)}{6}-5\cdot\frac{n(n+1)}{2}-n\right)=$$ $$=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}.$$

Michael Rozenberg
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It is interesting to note that $$\begin{align} \sum_{i=1}^n i^{\,4} &=\frac 1{30}n(n+1)(2n+1)(3n^2+3n-1)\\ &=\frac{3n^2+3n-1}5\cdot\frac {n(n+1)(2n+1)}6 \\ &=\frac{3n^2+3n-1}5\sum_{i=1}^n i^{\,2}\end{align}$$

Hypergeometricx
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yes there is a formula, $$\sum_{i=1}^ni^4=1/30\,n \left( 2\,n+1 \right) \left( n+1 \right) \left( 3\,{n}^{2}+3 \,n-1 \right) $$

Dr. Sonnhard Graubner
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