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Apologies to bother you with this, but how is the following arithmetic sequence solved?

$$\dfrac1n \left(\sum_{k=1}^{n-1}\dfrac{n-k+1}2\right)$$

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I do not understand the notation at all. Can you edit it into Latex format? – ncmathsadist Apr 24 '13 at 19:40
How to write math? For some basic information about writing math at this site see e.g. here, here, ... – Américo Tavares Apr 24 '13 at 19:42
Sorry guys this is my first time using this site. I will attempt to make this clearer – user74185 Apr 24 '13 at 19:46
The left side seems to mean $$\sum_{k=1}^n k.$$ What is the right side? – gt6989b Apr 24 '13 at 19:51
In my basic terms, this formula is 1/n multiplied by Sigma of n-k+1 (numerator) divided by 2 (denominator). Where the term below sigma is K=1 and the term above is N-1 – user74185 Apr 24 '13 at 20:07

First, you can pull out everything from the sum that does not depend on $k$. So $$\dfrac1n \left(\sum_{k=1}^{n-1}\dfrac{n-k+1}2\right)=\frac 1n \left(\frac {(n-1)(n+1)}2-\frac 12\sum_{k=1}^{n-1}k\right)$$ where I pulled out $\frac {n+1}2$ and multiplied by $n-1$ as the number of terms. Can you do the last?

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@P..: I see. Soon to fix – Ross Millikan Apr 24 '13 at 20:48
Apologies I still cannot use the correct math symbols, but I have this derivation as ((n-1)/2) - (1/2n) where this last term is multiplied by sigma K. – user74185 Apr 24 '13 at 20:56
Does this sound correct. I need the final answer only in terms on n, so what would happen to the lone K now??? – user74185 Apr 24 '13 at 20:57
@user74185: the $k$ is a dummy variable. It represents all the values you are summing up. Less formally, the last sum represents (without the leading $\frac 12$) $1+2+3+\ldots n-1$. There is no $k$. Have you seen that sum? You might look at the triangular numbers. – Ross Millikan Apr 24 '13 at 21:01
So we completely drop the k, which would lead to the final solution I derived from your formula, which was ((n-1)/2) - (1/2n) – user74185 Apr 24 '13 at 21:09

Write $$ \dfrac1n \left(\sum_{k=1}^{n-1}\dfrac{n-k+1}2\right)=\dfrac{n}{2n}\sum_{k=1}^{n-1}1-\dfrac1{2n}\sum_{k=1}^{n-1}k+\dfrac1{2n}\sum_{k=1}^{n-1}1 $$ and see this.

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