Question 2
Consider the sum of the three tables below. The first represents $n$ lines, the first with $n$ numbers equal to $n$, the second with $n-1$ numbers equal to $n-1$ And so on. In the second, we have a similar distribution of numbers, but in columns instead of rows. On the third, we have these numbers diagonals, the first diagonal has a number $1$, the second two numbers equal to $2$, the third three numbers equal to $3$ And so on.
The result of the sum of the three tables will be a table with the same number of numbers and with each position being the result of the sum of the corresponding positions in the three tables. For example, in the upper left corner, we will have the number $n+n+1=2n+1$.
a) One way to check how many numbers there are in each table is to turn one upside down and join with another to form a rectangle with $n$ rows and twice as many numbers in a table. Knowing this, how many numbers are there in a table?
b) How many times does each number appear? $k$ in all three tables?
c) For each position, line $i$ and column $j$, Determine the numbers written on it in the three tables and in the result table.
d) Using the information in the previous items, check that
$$
1^{2}+2^{2}+3^{2}+\ldots+(n-1)^{2}+n^{2}=\frac{1}{6} n(n+1)(2 n+1)
$$
The result of the sum of the three tables will be a table with the same number of numbers and with each position being the result of the sum of the corresponding positions in the three tables. For example, in the upper left corner, we will have the number $n+n+1=2n+1$.
a) One way to check how many numbers there are in each table is to turn one upside down and join with another to form a rectangle with $n$ rows and twice as many numbers in a table. Knowing this, how many numbers are there in a table?
b) How many times does each number appear? $k$ in all three tables?
c) For each position, line $i$ and column $j$, Determine the numbers written on it in the three tables and in the result table.
d) Using the information in the previous items, check that
$$
1^{2}+2^{2}+3^{2}+\ldots+(n-1)^{2}+n^{2}=\frac{1}{6} n(n+1)(2 n+1)
$$