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How can I write $1^2+3^2+\dots+(2n+3)^2$ without using the dots?

Please help me.

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closed as off-topic by user21820, José Carlos Santos, Did, Shailesh, YiFan Feb 6 at 0:43

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    $\begingroup$ Well, that you've generalize that $(2n+3)^2$ is a term should be a huge hint. $1^2 = (2*(-1) + 3)^2$ and $3=(2*0 + 3)^2$ and the next term is $5^2 = (2*1 + 3)^2$. So this is $\sum\limits_{k=-1}^n(2n+3)^2$. $\endgroup$ – fleablood May 10 '18 at 16:18
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    $\begingroup$ It's unclear whether you are asking about notation (which is how I read it, thus my tag edit) or about finding a closed-form formula for the series (which would imply that "summation" would be a good tag). This confusion is reflected in the different answers you have been given. $\endgroup$ – Joffan May 10 '18 at 17:00
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$$ \sum_{k=0}^{n+1}(2k+1)^2. $$Do you know the symbol $\sum$?

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  • $\begingroup$ Why not $2n+3$? $\endgroup$ – user557276 May 10 '18 at 16:27
  • $\begingroup$ @user557276 $2n+3$ is obtained as $2(n+1)+1$. $\endgroup$ – Przemysław Scherwentke May 10 '18 at 16:28
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$$\sum_{k=0}^{n+1} (2k+1)^2 = 1^2+3^2+\dots+(2n+3)^2.$$

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$$1^2+3^2+\dots+(2n+3)^2= (1^2+2^2+\dots+(2n+3)^2)-4(1^2+2^2+\dots+(n+1)^2)$$ $$= {(2n+3)(2n+4)(4n+7)\over 6}-4{(n+1)(n+2)(2n+3)\over 6}$$ $$= 2(n+2)(2n+3){4n+7-2n-2\over 6}$$ $$= (n+2)(2n+3){2n+15\over 3}$$ It is without dots :).

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\begin{align} 1^2+3^2+\dots+(2n+3)^2 &=1^2+2^2+3^2+\dots+(2n+2)^2+(2n+3)^2+(2n+4)^2\\ &\qquad-4(1^2+2^2+\dots+(n+2)^2)\\[6px] &=\frac{1}{3}(2n+4)\left(2n+4+\frac{1}{2}\right)(2n+4+1)\\ &\qquad-4\cdot\frac{1}{3}(n+2)\left(n+2+\frac{1}{2}\right)(n+2+1)\\[6px] &=\frac{2}{3}(n+2)\left(\left(2n+\frac{9}{2}\right)(2n+5)-2\left(n+\frac{5}{2}\right)(n+3)\right)\\[6px] &=\frac{1}{3}(n+2)(2n+5)(4n+9-2(n+3))\\[6px] &=\frac{1}{3}(n+2)(2n+3)(2n+5) \end{align}

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