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

Question

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

Please help me.

Answer 1

$$ \sum_{k=0}^{n+1}(2k+1)^2. $$Do you know the symbol $\sum$?

Answer 2

$$\sum_{k=0}^{n+1} (2k+1)^2 = 1^2+3^2+\dots+(2n+3)^2.$$

Answer 3

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

Answer 4

\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}