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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?

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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?2011-08-17
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    This is the simplest case of the arithmetic series, see http://en.wikipedia.org/wiki/Arithmetic_series for detailed explanations.2011-08-17
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    Also note the last term in your sum is $(a-n)=(a-(a-1))=1$2011-08-17
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    So my attempt was to simplify it to (n+1)a-n(n+1)/2 >= 1000, is that right?2011-08-17
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    It's not an equation; it's an inequality.2011-08-17
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    @aadidasu Yes, now presumably you want to find the smallest such n satIsfying that inequality. You should plug n+1 back in for a.2011-08-17
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    List the terms in your sequence, *concretely*, starting from the end. You will see what's going on. You don't *really* need $a$, sum is $(n+1)(n+2)/2$. Then calculate for a few values of $n$.2011-08-17

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