Consider the sets of consecutive integers {1}, {2, 3}, {4, 5, 6}, {7,

Mona2019 's solution is better.
The sets are:
S1 = (1)
S2 = (2,3)
S3 = (4,5,6)
S4 = (7,8,9,10)
S5 = (7,8,9,10,11)
....

The best part of the question is that it has patterns, however, they are the worst part as well. We can be sure that A, B and C can never be the answers(we know by reverse solving it).
Here, we can either try to figure out either the first element of the set or the last one.
I couldn't find a find way to find the first element so i tried the later and here's how i did.

Each set's last element if divided by that set number results in an increment of 0.5. For example -
For S(1), multiplication factor f(1) - \(\frac{1}{1} = 1\)
for S(2), f(2) = \(\frac{3}{2} = 1.5\)
for S(3), f(3) = \(\frac{6}{3} = 2.0\)
... so on

So, we now need to find number of increments when we reach the S(100). The increments are obtained by (n-1) where n is the last set upto which we are trying to figure out the increments.

Since there are 100 sets, number of increments = 100 - 1 = 99
Total increment = 99*0.5 = 49.5
Hence, f(50) = total increment + 1 = 49.5 + 1 = 50.5
the last element of S(100), x = 50.5*100 = 5050

Now, we need to find the first element of S(100), which is 5050 - 100 + 1 = 4951
Finally, we are trying to find \(S_100\) i.e. the sum of 4951, 4952, ... 5050.
The middle terms of S(100) are 5000 and 5001.
Thus, the S(100) = (5000+5001)/2*100 = 5000.5*100 = 500050

Answer E.

Surely, there must be an easier way.