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?

I will do it as the following,
$$
\begin{align}
&(1+n) + 1+(n-1) + 1+(n-2) + 1+(n-3) + 1+(n-4) + \cdots + 1+(n-n)\\
=&(1+n) + [1+(n-1) + 1+(n-2) + 1+(n-3) + 1+(n-4) + \cdots + 1+(n-n)]\\
=&(n+1) + [n+(n-1)+\cdots+2+1]\\
=&\frac{(n+1)[1+(n+1)]}{2}\\
=&\frac{(n+1)(n+2)}{2} 
\end{align}
$$
Now you can go on with Rolando's answer. 
