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I am practicing mathematical induction and I got this question. $$1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}$$

I am wondering, if this is correct way to do it? Would my answer get accepted if it was an exam?

I put n+1 in the places of n. So Im trying to get this final result:

$$\frac{(n+1)(n+2)(2n+3)}{6}$$

Getting the left side done: $$\frac{n(n+1)(2n+1)}{6} + \frac{6(n+1)^2}{6} = \frac{n(n+1)(2n+1) +6(n+1)^2}{6} = \frac{(n+1)(n+2)(2n+3)}{6}$$

Am I doing this completely wrong, can someone guide me?

Robert Shore
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etoRatio
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2 Answers2

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The proof is more or less correct, but you are missing some bureaucracy.

  1. You’re missing a begin of induction. Something like: For $n=0$ the property holds, because $0 = 0\cdot 1\cdot 1/6$
  2. You should explain a bit more in the step. Something like: From our induction hypothesis we know that $1^2+\ldots+n^2 = n(n+1)(2n+1)/6$. Thus $1^2+\ldots+(n+1)^2 = n(n+1)(2n+1)/6 + (n+1)^2 = \ldots$.

The first point is very important, as the proof is not correct without it, and the second one is at least relvant, because else other people might have problems understanding what you are actually doing.

Lazy
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It is quite well done, but you should check first that the statement is true when $n=1$, and then to say that if$$1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}6,\tag1$$then$$1^2+2^2+\cdots+n^2+(n+1)^2=\frac{n(n+1)(2n+1)}6+(n+1)^2$$(due to $(1)$), followed by the computations that you have shown us at the end of your question.

José Carlos Santos
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  • Do I need to simplify this $\frac{n(n+1)(2n+1)+6(n+1)^2}{6}$ or it's okay to write $\frac{(n+1)(n+2)(2n+3)}{6}$ after this? – etoRatio Sep 03 '21 at 22:14
  • It is probably more prudent to expand both $n(n+1)+6(n+1)^2$ and $(n+1)(n+2)(2n+3)$ and to remark that both expansions are equal to $2 n^3+9 n^2+13 n+6$; it follows from this that$$n(n+1)+6(n+1)^2=(n+1)(n+2)(2n+3).$$ – José Carlos Santos Sep 03 '21 at 22:22