The first number in each group: 1, 2, 5, 10, 17.....
Their common difference is in Arithmetic Progression. Hence, the general term of the series can be expressed as a quadratic equation.
Let the quadratic equation of the general term be $$ax^2+bx+c$$
1st term = a+b+c=1
2nd term = 4a+2b+c=2
3rd term = 9a+3b+c=5
Solving the equations, we get a=1, b=-2, c=2.
Hence, the first term in the 15th group will be $$\left(15\right)^2-2\left(15\right)+2=197$$
We can see that the number of terms in each group is 1, 3, 5, 7.... and so on. These are of the form 2n-1. Hence, the number of terms in 15th group will be 29. Hence, the last term in the 15th group will be 197+29-1 = 225.
Sum of terms in group 15= $$\frac{29}{2}\left(197+225\right)\ =\ 6119$$
Alternatively,
The final term in each group is the square of the group number.
In the first group 1, second group 4, ............
The final element of the 14th group is $$\left(14\right)^2=\ 196$$, similarly for the 15th group this is : $$\left(15\right)^2=\ 225$$
Each group contains all the consecutive elements in this range.
Hence the 15th group the elements are:
(197, 198, ................................225).
This is an Arithmetic Progression with a common difference of 1 and the number of element 29.
Hence the sum is given by : $$\frac{n}{2}\cdot\left(first\ term\ +last\ term\right)$$
$$\frac{29}{2}\cdot\left(197+225\right)$$
6119.
