Tenth Row?

$\begin{array} {ccc} \text{Row number }n & & \text{Row sum }S_n \\ \color{#D61F06}1 & 1=1 & {\color{#D61F06}1}^3 \\ \color{#D61F06}2 & 3+5=8 & {\color{#D61F06}2}^3 \\ \color{#D61F06}3 & 7+9+11=27 & {\color{#D61F06}3}^3 \\ \color{#D61F06}4 & 13+15+17+19 =64 & {\color{#D61F06}4}^3 \end{array}$

From the above we can deduce that the sum of the numbers in the $n$ th row $S_n = n^3$ , therefore that of the 10th row is $S_{10} = 10^3 = \boxed{1000}$ .

Let us prove the claim $S_n = n^3$ is true for all $n \in \mathbb N$ .

Proof:

We note that $\{a_k\}$ , the terms of $n$ th row, are in an arithmetic progression . From the first few $n$ , we note that the first term is given by $a_1 = n(n-1)+1=n^2-n+1$ . The common difference is $d=2$ and the number of term is $n$ . Therefore, the sum of the terms of $n$ th row is

$\begin{aligned} S_n & = \frac {n(2a_1+(n-1)d)}2 \\ & = \frac {n\left(2(n^2-n+1)+2(n-1)\right)}2 \\ & = n\left(n^2-n+1+n-1\right) \\ & = n^3 \ \square \end{aligned}$

Observe that the sum of the numbers in the $n^{th}$ row equals the sum of all of the numbers in the table minus the sum of the numbers from the first to the $(n - 1)^{th}$ row. For instance, for the fourth row we have

${\color{#3D99F6}{13 + 15 + 17 + 19}} \ \ {\large{=}} \ \ \begin{array} {c} {\color{#D61F06}{1}} \\ {\color{#D61F06}{+ \ 3 + 5}} \\ {\color{#D61F06}{+ \ 7 + 9 + 11}} \\ {\color{#3D99F6}{+ \ 13 + 15 + 17 + 19}} \end{array} \ \ {\large{-}} \ \ \begin{array} {c} {\color{#D61F06}{1}} \\ {\color{#D61F06}{+ \ 3 + 5}} \\ {\color{#D61F06}{+ \ 7 + 9 + 11}} \end{array}$

The sum of all of the numbers in the table is the sum of the first $n$ odd numbers, which is $n^2$ . We need a way to find the $n^{\text{th}}$ odd number - that is, the last number on each row - so that we can sum up the first $n$ numbers on the table. Let $O_n$ denote the last number on the $n^{\text{th}}$ row, and $T_n$ the $n^{\text{th}}$ triangular number. We observe that $O_n = T_n$ . For instance,

$\begin{array} {c} {\color{#3D99F6}{1}} & \small {\color{#D61F06}{1^{\text{st}}}} \ \text{row} \implies \ \text{The} \ {\color{#D61F06}{\text{first}}} \ \text{triangular number is} \ \ {\color{#20A900}{1}} \implies \ {\color{#3D99F6}{1}} \ \text{is the} \ {\color{#20A900}{\text{first}}} \ \text{odd number} \\ 3 + {\color{#3D99F6}{5}} & \small {\color{#D61F06} 2^{\text{nd}}} \ \text{row} \ \implies \ \text{The} \ {\color{#D61F06}{\text{second}}} \ \text{triangular number is} \ {\color{#20A900}{3}} \ \implies \ {\color{#3D99F6}{5}} \ \text{is the} \ {\color{#20A900}{\text{third}}} \ \text{odd number} \\ 7 + 9 + {\color{#3D99F6}{11}} & \small {\color{#D61F06}{3^{\text{rd}}}} \ \text{row} \implies \ \text{The} \ {\color{#D61F06}{\text{third}}} \ \text{triangular number is} \ \ {\color{#20A900}{6}} \ \implies \ {\color{#3D99F6}{11}} \ \text{is the} \ {\color{#20A900}{\text{sixth}}} \ \text{odd number} \\ 13 + 15 + 17 + {\color{#3D99F6}{19}} & \small {\color{#D61F06}{4^{\text{th}}}} \ \text{row} \implies \ \text{The} \ {\color{#D61F06}{\text{fourth}}} \ \text{triangular number is} \ \ {\color{#20A900}{10}} \ \implies \ {\color{#3D99F6}{19}} \ \text{is the} \ {\color{#20A900}{\text{tenth}}} \ \text{odd number} \end{array}$

Since the sum of the first $n$ odd numbers is $n^2$ , then $\left(O_n\right)^2 = \left(T_n\right)^2$ .

Let the sum of the numbers in the tenth row be $S_{10}$ . Combining our first and second observations, we have

$\displaystyle \begin{aligned} S_{10} & = \left(O_{10}\right)^2 - \left(O_9\right)^2 \\ & = \left(T_{10}\right)^2 - \left(T_9\right)^2 & \small \color{#3D99F6} T_{10} = \frac{10 \cdot 11}{2} = 55 \ , \ T_9 = \frac{9 \cdot 10}{2} = 45 \\ & = 55^2 - 45^2 \\ & = (55 - 45)(55 + 45) \\ & = 10 \cdot 100 \\ & = \boxed{10000} \end{aligned}$