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Have I started this right? I know I have to add $(k+1)$ but why?

Use mathematical induction to prove that $$1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}.$$

BASIS STEP: $P(0)$ is true since $$\begin{align}(2(0)+1)^2 &= \frac{(0+1)(2(0)+1)(2(0)+3)}{3}\\1 &= \frac{3}{3}\\1 &= 1.\end{align}$$

INDUCTIVE STEP: Assume $P(k)$ holds for an arbitrary integer $k>0$. $$1^2+ 3^2 + 5^2+ \cdots +(2k + 1)^2=\frac{(k+1)(2k+1)(2k + 3)}{3}$$

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    $\begingroup$ Please, use $\LaTeX$. $\endgroup$ – João Jun 17 '13 at 0:25
  • $\begingroup$ Thank you for not helping at all. If I knew Latex, don't you think I would use it? $\endgroup$ – Hermes Trismegistus Jun 17 '13 at 0:26
  • $\begingroup$ You want to prove that the sum of the $(2n-1)^{th}$ perfect squares is equal to $\frac{(n+1)(2n+1)(2n+3)}{3}$? $\endgroup$ – João Jun 17 '13 at 0:31
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You've proved that the statement is true for $k=0$ and now supposing that is true to an arbitrary $k$,

$$1^2+2^2+\cdots+(2k+1)^2 = \frac{(k+1)(2k+1)(2k+3)}{3}$$ let's see the case $k+1$. Then we use the hypothesis of induction and we get $$\begin{align}1^2+2^2+\cdots+(2k+1)^2+(2(k+1)+1)^2 &= \frac{(k+1)(2k+1)(2k+3)}{3}+(2(k+1)+1)^2\\ &= \frac{(k+1)(2k+1)(2k+3)+3(2k+3)^2}{3}\\ &= \frac{[(k+1)(2k+1)+3(2k+3)](2k+3)}{3}\\ &= \frac{[2k^2+3k+1+6k+9](2k+3)}{3}\\ &= \frac{(2k+5)(k+2)(2k+3)}{3}\\ &= \frac{[(k+1)+1][2(k+1)+1][2(k+1)+3]}{3},\end{align}$$ where we used that $(2k+5)(k+2)=2k^2+9k+10.$

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You need to assume that it holds for some $k\ge0,$ not some $k>0$, for otherwise the induction doesn't go through (and you've only proved it for $k=0$).

Now, given the assumption that $$1^2+ 3^2 + 5^2+ \cdots +(2k + 1)^2=\frac{(k+1)(2k+1)(2k + 3)}{3},$$ you need to add $\bigl(2(k+1)+1\bigr)^2$ to both sides, and show that that makes the right hand side equal to $$\frac{\bigl((k+1)+1\bigr)\bigl(2(k+1)+1\bigr)\bigl(2(k+1)+3\bigr)}3.$$ That's how you show inductively that $P(k+1)$ holds whenever $P(k)$ does.


As an alternative, show that $$0^2+1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6},$$ so that $$0^2+2^2+4^2+\cdots+(2k)^2=4(0^2+1^2+2^2+\cdot+k^2)=\frac{2k(k+1)(2k+1)}{3}$$ and $$\begin{align}0^2+1^2+2^2+\cdots+(2k+1)^2 &= \frac{(2k+1)\bigl((2k+1)+1\bigr)\bigl(2(2k+1)+1\bigr)}{6}\\ &= \frac{(2k+1)(2k+2)(4k+3)}{6}\\ &= \frac{(k+1)(2k+1)(4k+3)}{3}.\end{align}$$ Can you see how to proceed from there?

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