4 & 5 & 6 & 7 \\ \hline
5 & 6 & 7 & 8 \\ \hline
6 & 7 & 8 & 9 \\ \hline
7 & 8 & 9 & 10 \\ \hline
\end{array}$ (b) $14\ 652$ (c) $1\ 930\ 797$ (d) No
- The required table
$\begin{array}{|c|c|c|c|} \hline
4 & 5 & 6 & 7 \\ \hline
5 & 6 & 7 & 8 \\ \hline
6 & 7 & 8 & 9 \\ \hline
7 & 8 & 9 & 10 \\ \hline
\end{array}$ - Note that the first number and the last number of the 1st row of the 99th table are $99$ and $197$, and there are $99$ numbers in the 1st row. Hence, the required sum
$\begin{array}{cl}
= & \dfrac{(99+197)\times 99}{2} \\
= & 14~652
\end{array}$ - Note that the sums of all integers in the rows form an arithmetics sequence. Note also that the sum of all integers in the 2nd row is larger than that of the 1st row by $99$. Therefore, the common difference is $99$. There are $99$ rows in the 99th table, then there are $99$ terms in the sequence. Hence, the required sum
$\begin{array}{cl}
= & \dfrac{99}{2} [ 2(14~652) +(99-1)(99)] \\
= & 1~930~797
\end{array}$ - For an odd number $k$, the sum of all integers in the 1st row of the $k$th table
$\begin{array}{cl}
= & \dfrac{k}{2}[2k+(k-1)(1)] \\
= & \dfrac{k(3k-1)}{2}
\end{array}$Note that the sums of all integers in the rows form an arithmetics sequence, and the common difference is $k$. Since there are $k$ rows in the $k$th table, then there are $k$ terms in the sequence. Hence, the sum of all integers
$\begin{array}{cl}
= & \dfrac{k}{2}\left[ 2 \times \dfrac{k(3k-1)}{2} +(k-1)k\right] \\
= & \dfrac{k}{2} (4k^2 -2k) \\
= & k^2(2k-1)
\end{array}$Since $k$ is an odd number, then $k^2$ and $(2k-1)$ are also odd numbers. Since the product of two odd numbers are odd, then $k^2(2k-1)$ is also an odd number. Therefore, there is no odd number $k$ such that the sum of all integers in the $k$th table is an even number.
