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?