Comes back my odd love!

The pattern goes on as follows:
The sum of the number in the first bracket: $1$
The sum of the numbers in the second bracket: $9$
The sum of the numbers in the third bracket: $35$
The sum of the numbers in the fourth bracket: $91$
...........
So the pattern is $1,9,35,91,\dots$ Let n be the number of the bracket.
So the sum of the numbers in any bracket can be calculated as ${(n)}^3+{(n-1)}^3.$
eg.The sum of numbers in the $3^{rd}$ bracket $=3^3+2^3=35.$
Therefore the sum of the numbers in the ${23}^{rd}$ bracket $={23}^3+{22}^3=\boxed{22815}.$

We note that the number of terms or numbers in $k$ th bracket is given by $n_k =2k-1$ . For $k=23$ , $n_{23} = 2\cdot 23-1= 45$ .

We also note that the last term of $k$ th bracket is the sum of numbers of terms from first to $k$ th brackets inclusively. That is, $l_k = \dfrac {n_k(n_k+1)}2$ . For $k = 23$ , $l_{23} = \dfrac {n_{23}\left(n_{23}+1\right)}2 = \dfrac{45(45+1)}2 = 529$ . The last term of 22nd bracket is $l_{22} = \dfrac {43(43+1)}2 = 484$ , therefore, the first term of the 23rd bracket is $l_{22}+1 = 485$ . Therefore, the sum of the numbers in the 23rd bracket is:

$\begin{aligned} S_{23} & = \frac {n_{23}\left(a_{23}+l_{23} \right)}2 \\ & = \frac {45(485+529)}2 \\ & = \boxed{22815} \end{aligned}$

Notice that, the numbers in each bracket are in $\color{#20A900}{\text{Arithmetic Progression}}$ .

Also, notice that, first term, last term and number of terms in $n^{th}$ bracket is $n^{2}-2n+2$ , $n^{2}$ and $2n-1$ respectively.

Using the fact that the sum of terms in an $\color{#20A900}{\text{Arithmetic Progression}}$ that consists of $n$ terms such that first term is $a$ and last term is $b$ is given by $\dfrac{n}{2} \,\cdot \, (a+b)$ , we can write :-

$\huge \boxed{\color{#EC7300}{S_n\,=\,(2n-1)\cdot (n^{2}-n+1)}}$ .

,where $S_n$ denotes the sum of numbers present in the $n^{th}$ bracket.

On putting $n=23$ , $S_{23}$ comes to be equal to $\boxed{22815}$ .