How can I use induction to prove that
$1^2+3^2+...+(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ for non-negative integers $n$?
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What did you try? – José Carlos Santos May 05 '18 at 09:20
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1I would start with $n=1$ – Dr. Sonnhard Graubner May 05 '18 at 09:21
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@JoséCarlosSantos Nothing because I don't know where to start. – user557276 May 05 '18 at 09:21
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1If you don't know that you should start by checking whether it holds or not when $n=1$, then perhaps that you should learn about induction from the start. – José Carlos Santos May 05 '18 at 09:22
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1Every statement of the form $f(0)+f(1)+\cdots+f(n)=g(n)$ can be proved by verifying two things: (i) $g(0)=f(0)$ and (ii) $g(n+1)-g(n)=f(n+1)$. – Angina Seng May 05 '18 at 09:26
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Try to follow the given hints and show your work here, we'll take a look to it! You can also check on MSE network for similar questions to have a guide. Good work! – user May 05 '18 at 09:33
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Take a look here for example https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct – user May 05 '18 at 09:35
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@user557276 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user May 31 '18 at 20:02
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HINT
Recall that proof by induction require two steps
1. Base case
- we need to check that the statement is true for $n=1$
2. Induction step
- assume $1^2+3^2+...+(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ as hypotesis and by that try to prove that the following holds
$$1^2+3^2+...+(2n+1)^2+2[(n+1)+1]^2=\frac{[(n+1)+1][2(n+1)+1][2(n+1)+3]}{3}$$
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