Set of sets

We see that the last term of $S_n$ is the $n^{th}$ triangular number $=\dfrac{n(n+1)}{2}$

So , last term of $S_{50}=\dfrac{50(50+1)}{2}=25\times 51=1275$

Since , the last term is $1275$ and there are $50$ terms in the set , the first term is $1226$ .

We see that the numbers of $S_n$ form an Arithmetic progression and thus their sum $=\dfrac{n}{2}[2a+(n-1)d]$

Sum of members of $S_{50} = \dfrac{50}{2}[2(1226)+(50-1)1] = 61300+1225 = \huge\boxed{\color{#3D99F6}{62525}}$

Let the members of $\space S_n = \{ a_{n,1}, a_{n,2},a_{n,3},...,a_{n,n} \}$ . We note that the $i^{th}$ member of $S_n$ is given by:

$a_{n,i} = \displaystyle \sum_{k=1}^{n-1} {k} + i = \dfrac {(n-1)n}{2} + i$

Therefore, the sum of the members of $S_{50}$ is:

$\begin{aligned} \sum_{k=1}^{50} {a_{50,k}} & = \dfrac {50(a_{50,1} + a_{50,50})}{2} = 25\left( 2 \times \dfrac {49\times 50}{2} + 1 + 50 \right) \\ & = 25(49\times 50 + 51) = \boxed{62525} \end{aligned}$

Nice question.

sum = n cubed/2 + n/2