A problem from Observations on Arithmetica In his Observations on Diophantus' Arithmetica, Fermat writes the following (translated)

In the natural progression starting at unity, the product of an arbitrary number times its immediate successor makes double the triangle of the first number.
If the multiplier is the triangle of the number immediately following, we have three times the pyramid of the first number.
If it is the pyramid of the number immediately following, we have the quadruple of the “triangulo-triangulaire” of the first number, and so on indefinitely, following a uniform
and general rule.

Question 1

What would this be in modern symbolic notation?

Question 2

What would be a proof for the same?

Any help will be appreciated.
Thanks.
 A: Let $T_n$ denote the $n$th triangular number, i.e. $\sum_{i=1}^n i$, and let $T^{(2)}_n$ denote the $n$th pyramidal number, i.e. $\sum_{i=1}^n T_i$. Then the first sentence says:

$n(n+1)=2T_n$

This is simply the usual formula for $T_n$, which can be proved by induction. We usually write it:
$$T_n=\frac{n(n+1)}{2}=\binom{n+1}{2}$$
The second sentence says:

$nT_{n+1}=3T^{(2)}_n$

We can also show this by induction, using the previous formula. In notation more in line with modern sensibilities:
$$T^{(2)}_n = \frac{nT_{n+1}}{3} = \frac{n(n+1)(n+2)}{6}=\binom{n+2}{3}$$
The last one, I don't know a modern name for, but let's let $T^{(3)}_n=\sum_{i=1}^nT^{(2)}_i$. Fermat seems to be telling us that:

$nT^{(2)}_{n+1}=4T^{(3)}_n$

We would write this today as:
$$T^{(3)}_n=\frac{nT^{(2)}_{n+1}}{4} = \frac{n(n+1)(n+2)(n+3)}{24}=\binom{n+3}{4}$$
I'm sure we could prove this by induction, again using the above formula.
Finally, Fermat tells us that this pattern continues. In other words:
$$T^{(k)}_n=\frac{nT^{(k-1)}_{n+1}}{k+1} = \frac{(n+k)!}{(n-1)!(k+1)!}=\binom{n+k}{k+1}$$
I reckon this proof would be a double-induction.
