Determine the sum of the last three elements of a line of the arithmetic triangle, knowing that the third element of the line before that is equal to $ 79800 $ and that the sum of the first three elements of this previous line is equal to $ 80201 $
Could anyone give me any tips for this problem?
You know that the third element of the line before is equal to $79800$? But that is nothing else than the binomial coefficient $79800 = {n \choose 2}$ for some $n\in \mathbb{N}$.
You know as well that the sum of the first three elements in any line of the arithmetic triangle is
$${n \choose 0} + {n \choose 1} + {n\choose 2} = 1 + n + {n \choose 2}$$.
As you already know that ${n\choose 2} = 79800$ we can calculate $n$ to be:
$n = 80201 - 1 -{ n \choose 2} = 80200 - 79800 = 400$.
Thus the next line has $n=401$ and the sum of the last three elements is just
$${401 \choose 399} + {401 \choose 400} + {401 \choose 401} = \frac{401\cdot 400}{2} + 401 + 1 = 80602.$$