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.
The problem said "by writing the summation in expanded form". I would take that to mean
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:
.
So the problem reduces to finding two sums: and . Whoever gave you this problem expects you to know simple formulas for those sums.