# Thread: help understanding summation equivalance

1. ## help understanding summation equivalance

could someone break this down into intermediate steps for me, i realy am not seeing the pattern here.

2. ## Re: help understanding summation equivalance

Prove \displaystyle \begin{align*} \sum_{j = 1}^k{ \left[ \frac{2}{k} \left( k - j + 1 \right) \right] } = \frac{2}{k}\left[ k + \left( k - 1 \right) + \left( k - 2 \right) + \dots + 3 + 2 + 1 \right] \end{align*}

No pattern is needed. This is the intermediate step, which is using j as your counter and plugging in. First note that the factor of \displaystyle \begin{align*} \frac{2}{k} \end{align*} is constant, so it can be pulled out of the sum. Then

\displaystyle \begin{align*} j = 1 \end{align*} gives \displaystyle \begin{align*} k - 1 + 1 = k \end{align*}

\displaystyle \begin{align*} j = 2 \end{align*} gives \displaystyle \begin{align*} k - 2 + 1 = k - 1 \end{align*}

\displaystyle \begin{align*} j = 3 \end{align*} gives \displaystyle \begin{align*} k - 3 + 1 = k - 2 \end{align*}

\displaystyle \begin{align*} j = 4 \end{align*} gives \displaystyle \begin{align*} k - 4 + 1 = k - 3 \end{align*}

\displaystyle \begin{align*} \vdots \end{align*}

\displaystyle \begin{align*} j = k - 2 \end{align*} gives \displaystyle \begin{align*} k - \left( k - 2 \right) + 1 = 3 \end{align*}

\displaystyle \begin{align*} j = k - 1 \end{align*} gives \displaystyle \begin{align*} k - \left( k - 1 \right) + 1 = 2 \end{align*}

\displaystyle \begin{align*} j = k \end{align*} gives \displaystyle \begin{align*} k - k + 1 = 1 \end{align*}

so adding up all these terms gives

\displaystyle \begin{align*} k + \left( k - 1 \right) + \left( k - 2 \right) + \left( k - 3 \right) + \dots + 3 + 2 + 1 \end{align*}.

Also, this simplifies a lot more as it is an arithmetic series.

3. ## Re: help understanding summation equivalance

The problem said "by writing the summation in expanded form". I would take that to mean
$\sum_{j=1}^k \frac{1}{k}(2k- j+ 1)= \sum\frac{1}{k}(2k)- \sum_{j=1}^k j+ \sum_{j=1}^k 1$

Do you understand that "k" is a constant and j is the index of summation? That means we can "factor" k out of the first sum:
$\frac{2k}{k}\sum_{j=1}^k 1- \frac{1}{k}\sum_{j=1}^k j+ \frac{1}{k}\sum_{j=1}^k 1$

$2\sum_{j=1}^k 1- \frac{1}{k}\sum_{j=1}^k j+ \frac{1}{k}\sum_{j=1}^k 1$.

So the problem reduces to finding two sums: $\sum_{j=1}^k 1= 1+ 1+ 1+\cdot\cdot\cdot+ 1$ and $\sum_{j=1}^k j= 1+ 2+ 3+\cdot\cdot\cdot+ k-1+ k$. Whoever gave you this problem expects you to know simple formulas for those sums.