I have an equation as:
$\sum_{r=1}^{n}n-r$
When I evaluate it, I get the result as: = (n-1)+(n-2)+..+(n-n) = ?
How to evaluate $\sum_{r=1}^{n}n-r$
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summation
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1Add up the "$n$" terms first. Then add up the subtracted terms. Then subtract. – 2017-02-13
2 Answers
3
Hint: $\sum_{r = 1}^n (n-r) = n^2 - \sum_{r=1}^n r$. Now you should be able to give a close form of $\sum_{r=1}^n r$.
3
Hint: \begin{eqnarray*} \sum_{r=1}^n n-r &=& (n-1)+(n-2)+\cdots + 1+ 0 \\ &=& 0 + 1 + \cdots +(n-2)+(n-1) \\ &=& \sum_{r=1}^{n-1}r.\end{eqnarray*}
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0So, that is nC2 – 2017-02-13
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1yep, ${n\choose 2}$. – 2017-02-13