I am not sure how to solve this equation. Any ideas
$$(1+n) + 1+(n-1) + 1+(n-2) + 1+(n-3) + 1+(n-4) + \cdots + 1+(n-n) \ge 1000$$ Assuming $1+n = a$ The equation can be made to looks like
$$a+(a-1)+(a-2)+(a-3)+(a-4)+\cdots+(a-n) \ge 1000$$
How to proceed ahead, or is there another approach to solve this?