Solve this Inequality 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?

Here is an example (probably not the most elegant though...) way to solve this. First either you write $(n+1)a-\frac{n(n+1)}{2}$ and replace $a$ by $n+1$ or you just realize replacing $a$ by $n+1$ in your second inequation that the left side is $\sum_{i=1}^{n+1}i=\frac{(n+1)(n+2)}{2}$.
Your inequality is then 
$$(n+1)^2+(n+1)\geq 2000$$
You can for example consider the polynomial $P(X)=X^2+X-2000$ and you want to study the sign of this, which should not be a problem.
