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

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Did, Shailesh
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• 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 $\sum\limits_{k=-1}^n(2n+3)^2$. – fleablood May 10 '18 at 16:18
• 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. – Joffan May 10 '18 at 17:00

$$\sum_{k=0}^{n+1}(2k+1)^2.$$Do you know the symbol $\sum$?
• Why not $2n+3$? – user557276 May 10 '18 at 16:27
• @user557276 $2n+3$ is obtained as $2(n+1)+1$. – Przemysław Scherwentke May 10 '18 at 16:28
$$\sum_{k=0}^{n+1} (2k+1)^2 = 1^2+3^2+\dots+(2n+3)^2.$$
$$1^2+3^2+\dots+(2n+3)^2= (1^2+2^2+\dots+(2n+3)^2)-4(1^2+2^2+\dots+(n+1)^2)$$ $$= {(2n+3)(2n+4)(4n+7)\over 6}-4{(n+1)(n+2)(2n+3)\over 6}$$ $$= 2(n+2)(2n+3){4n+7-2n-2\over 6}$$ $$= (n+2)(2n+3){2n+15\over 3}$$ It is without dots :).