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Prove by mathematical induction that $$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$ holds $\forall n\in\mathbb{N}$.

(1) Assume that $n=1$. Then left side is $1^2 =1$ and right side is $6/6 = 1$, so both sided are equal and expression holds for $n = 1$.

(2) Let $k \in \mathbb{N}$ is given. Assume that for $n = k$ expression holds. Then for $n = k+1$ we get $$\sum_{i = 1}^{k+1} i^2 = \left(\sum_{i = 1}^{k} i^2\right) + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + k^2 + 2k + 1 = \frac{2k^3 + 9k^2 + 13k + 6}{6}.$$ Factoring the result we get that $\frac{2k^3 + 9k^2 + 13k + 6}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$ and thus expression holds for $n = k+1$.

Combining (1) and (2) we can conclude that the expression holds $\forall n \in \mathbb{N}$.

I have a few questions:

  1. Is my proof correct?
  2. If you would be a math professor, is this style of writing math proofs right and sufficient for freshman? Or is there something I miss?
Accelerate to the Infinity
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    (1) Yes, completely. (2) I am, and I think your proof, as is written, is not only enough but in fact neater and clearer than what I many times see. – DonAntonio Sep 19 '16 at 15:32
  • @DonAntonio thanks. It was just, that they didn't teach us math induction at our HS, and our uni requires knowledge of it. So I was just asking, if I'm not missing anything. :-) – Accelerate to the Infinity Sep 19 '16 at 15:34
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    Minor corrections: Instead of "Assume that $n=1$" I would say "If $n=1$ then" or similar. It sound strange to use "assume" here. – Winther Sep 19 '16 at 15:37
  • Try proving that $\sum_{k=1}^nk(k+1)(k+2)\dots(k+p)=\frac{n(n+1)(n+2)\dots(n+p)(n+p+1)}{p+2}$ by induction. – Simply Beautiful Art Nov 24 '16 at 13:34

2 Answers2

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we have to show that $$\frac{n(n+1)(2n+1)}{6}+(n+1)^2=\frac{(n+1)(n+2)(2(n+1)+1)}{6}$$ this is true since $$\frac{n(n+1)(2n+1)}{6}+(n+1)^2$$ is the same as the right-hand side after some algebra

Dr. Sonnhard Graubner
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I believe that my answer is implicit in previous, but I tell you that since the factor $(k+1)$ should be in your final formula (I say the factor $(n+1)$ in main formula), you know it since your proof is by induction, and such factor was in each summand $\frac{k(k+1)(2k+1)}{6}$ and $(k+1)^2$ then I believe that is better strategy to calculate this sum as $$(k+1)\cdot\text{something}$$ and obtain your result factoring $\text{something}$, since is a polynomial in $k$ of degree $2$. Notice that your factorisation requires solve a polynomial of degree three. Thus your proof is right, and you can think about this detail if you think that could be useful next time (I am saying that such factor is a guide to get the final statement).