ld compare the numerators not based on r, the numbers in the row, but on n, the row number. To do this, we would plug the numbers into a table:
             The numerator series are: 1, 3, 6, 10, 15
             Now, I'll try to observe a pattern in these numerator series. From the given data, it is apparent that the numerator series is continuously increasing. If I plug these numbers into a table.
             From this, we can clearly see that the numerator for the next row (Row number = 6) will be 15 + 6 because the difference in the numerator series is progressively increasing by 1, so our numerator for the sixth row will be 21.
             From the above table, we can see that as the row number increases, the numerator is also increasing, but not at the same pace. The pace with which numerator is increasing is greater than the pace of increase in the row number. So, let us calculate the rate at which numerator is increasing w.r.t row number, by dividing the numerator is each row by its corresponding row number. Let us denote each of this rate by z.
             It is quite apparent from the above table that if we fit a simple linear regression graph between z and n, we would get a perfectly straight line because the consecutive difference between the elements (row number and z) is constant.
             Now, let us plot a simple linear regression line between z and n.
             So, from the graph we can see that an equation of the form z = 0.5n + 0.5 fits into these points.
             Numerator = n x (0.5n + 0.5) = = 0.5n2 + 0.5n
             Thus, general expression for getting the numerator of row 'n' = = 0.5n2 + 0.5n
             So, when n = 6, Numerator of 6th row = 0.5 (6)2 + 0.5 x 6 = 21
             And, when n = 7, Numerator of 7th row = 0.5 (7)2 + 0.5 x 7 = 28
             Let us represent the numerator of each row 'n' by N(n).
             ...