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I wonder if it is possible to compose formulas by hand like: $1+2+3+\cdots+n$. The formula for this is:$(n)(n+1)/2$.

And for $1^2 + 2^2 + 3^2+\cdots+n^2$ you have the formula: ...

But this is more a trial and error work. So my question is, is it possible too generate series of the form: $1^3+2^3+3^3+\cdots+n^3$ or $1^4+2^4+3^4+\cdots+n^4$, etc. by logic and not by trial and error?

Thank you

Brian Tung
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SavantCode
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1 Answers1

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The formula for $1^2 + 2^2 + 3^2+\cdots+n^2$ is not by trial and error. There is a proof by logic, which is as follows.

Note $$(n+1)^3-1^3=\sum_{k=1}^{n}(k+1)^3-k^3=\sum_{k=1}^{n}(3k^2+3k+1)$$ $$=3\sum_{k=1}^{n} k^2+3\sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$$

This gives us that $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3 + 3n^2 + n}{6}$$ We can do something similar for $n^3$, $n^4$, and there is a formuala for the $p$-th powers called Faulhaber's Formula.

S.C.B.
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  • Ah, thank you. Very nice – SavantCode Aug 10 '16 at 04:52
  • @NIelsSavant Out of curiosity, was this not in your book? This proof is mentioned in the second page of my high school textbook on sums. – S.C.B. Aug 10 '16 at 04:55
  • No. Realy I studied by myself programming and physics. And ofcourse I learned sums and series. But I never heard about a general formula to compute a serie of the n-th power – SavantCode Aug 10 '16 at 16:20