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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? Also, is there any name for this kind of a series. Will update the question with specific name if there is any

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Luckily, addition is associative and commutative. Get rid of the parentheses and try to do some cancelling. Hint: what is the sum 1 + 2 +...+n? – Joe Aug 17 '11 at 17:30
This is the simplest case of the arithmetic series, see for detailed explanations. – Did Aug 17 '11 at 17:32
Also note the last term in your sum is $(a-n)=(a-(a-1))=1$ – RHP Aug 17 '11 at 17:33
So my attempt was to simplify it to (n+1)a-n(n+1)/2 >= 1000, is that right? – aadidasu Aug 17 '11 at 17:37
It's not an equation; it's an inequality. – Michael Hardy Aug 17 '11 at 17:38

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.

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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.

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