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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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closed as off-topic by José Carlos Santos, Arjang, JMP, Karn Watcharasupat, TheGeekGreek May 5 '18 at 9:55

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  • $\begingroup$ What did you try? $\endgroup$ – José Carlos Santos May 5 '18 at 9:20
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    $\begingroup$ I would start with $n=1$ $\endgroup$ – Dr. Sonnhard Graubner May 5 '18 at 9:21
  • $\begingroup$ @JoséCarlosSantos Nothing because I don't know where to start. $\endgroup$ – user557276 May 5 '18 at 9:21
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    $\begingroup$ If 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. $\endgroup$ – José Carlos Santos May 5 '18 at 9:22
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    $\begingroup$ Every 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) $g(n+1)-g(n)=f(n+1)$. $\endgroup$ – Lord Shark the Unknown May 5 '18 at 9:26
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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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