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I do know formula for finding sum of squares of n odd numbers but I am not sure how to find closed formula.

Note: From closed formula I mean a reduced fraction in factored form, with no ellipses (“. . .”) in it.

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  • $\begingroup$ You you mean sum of squares of odd numbers, or sums of odd numbers, of which you have $n^2$ of them? $\endgroup$ – law-of-fives May 8 '17 at 4:13
  • $\begingroup$ Please try to place your actual question in both the body and the title... $\endgroup$ – fleablood May 8 '17 at 4:13
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Assuming you have the formulas \begin{align*} &\sum_{k=1}^n k = \frac{n(n+1)}{2}\\[4pt] &\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\\[4pt] \end{align*}

Then

\begin{align*} \sum_{k=1}^n (2k-1)^2 &= \sum_{k=1}^n 4k^2- 4k + 1\\[4pt] &= 4\sum_{k=1}^n k^2 -4\sum_{k=1}^n k +\sum_{k=1}^n 1 \\[4pt] &= 4\left(\frac{n(n+1)(2n+1)}{6}\right) -4\left(\frac{n(n+1)}{2}\right) +n\\[4pt] &=\frac{n(2n-1)(2n+1)}{3}\\[4pt] \end{align*}

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  • $\begingroup$ Which formula are you talking about? $\endgroup$ – Andre May 8 '17 at 4:28
  • $\begingroup$ Do you know the formulas $$1 + 2 + 3 + \cdots +\,n = \frac{n(n+1)}{2}$$ $$1^2 + 2^2 + 3^2 + \cdots +\,n^2 = \frac{n(n+1)(2n+1)}{6}$$ ?? $\endgroup$ – quasi May 8 '17 at 4:30
  • $\begingroup$ Yes I do know the formulas. Sorry I got confused. $\endgroup$ – Andre May 8 '17 at 6:08

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