A strange sequence

Another way to write this is:

$a_n = a_{n-1} + \displaystyle \sum_{x=1}^{n-1} x, \; n \geq 2,\; a_1 = 0$

$x=a_7\\ =a_6 + \displaystyle \sum_{x=1}^6 x\\ =35+\dfrac{6(7)}{2}\\ =35+21\\ =56$

$y=a_8\\ =a_7 + \displaystyle \sum_{x=1}^7 x\\ =56+\dfrac{7(8)}{2}\\ =56+28\\ =84$

$x \times y = 56 \times 84 = \boxed{4704}$

The sequence follows the rule: $a_n=\frac{n\cdot (n+1)\cdot(n+2)}{6}$ So the missing terms are 56 and 84, 56*84=4704

Another way could be to find the differences between two consecutives terms, they are $1, 3, 6, 10, 15...$ that are the sums from 1 to n

Conaecutively, a triangular number is being added to the next term. So, the first unknown term is $35 + 21 = 56$ and the next one is $56 + 28 = 84$ .
So, $x × y = 56 × 84 = \color{#69047E}{\boxed{4704}}$ .

I think the easiest way is finding the difference:

0, 1, 4, 10, 20, 35, $x$ , $y$

Difference:

(1 - 0), (4 - 1), (10 - 4), (20 - 10), (35 - 20)

= 1, 3, 6, 10, 15

Next 2 terms are:

1, 3, 6, 10, 15, 21 , 28

21 + 35 = 56

56 + 28 = 84

$x$ ( $y$ ) = 56 (84) = 4704