Imagine a triangle of numbers which follows this pattern:

 * Starting with the number "1", "1" is positioned at the top of the triangle. As this is the 1st row, it can only support a single number.
 * The 2nd row can support the next 2 numbers: "2" and "3"
 * Likewise, the 3rd row, can only support the next 3 numbers: "4", "5", "6"
 * And so on; this pattern continues.

```
    1
   2 3
  4 5 6
 7 8 9 10
...
```

Given N, return the sum of all numbers on the Nth Row:

1 <= N <= 10,000

Source: codewars | Difficulty: EASY
Tags: Geometry

Starter code: def cumulative_triangle(n):
	

Solutions:
def cumulative_triangle(n):
	return n * (n * n + 1) / 2


def cumulative_triangle(n):
	return (n ** 3 + n) / 2


cumulative_triangle = lambda n: (n ** 3 + n) / 2


def cumulative_triangle(n):
	start = sum(range(n)) + 1
	return sum(range(start, start + n))


def cumulative_triangle(n):
	f = lambda x: 0.5 * x * (x - 1) + 1
	s = f(n)
	r = []
	while len(r) < n:
		r.append(s)
		s += 1
	return sum(r)


cumulative_triangle = lambda n: n * (n * n + 1) / 2


def cumulative_triangle(n):
	x = sum(range(n + 1))
	return sum(range(x - n + 1, x + 1))


def cumulative_triangle(n):
	N = n * (n + 1) // 2
	prev = n * (n - 1) // 2
	return N * (N + 1) // 2 - prev * (prev + 1) // 2