Answer
$$\sum_{k=1}^{6} (-1)^{k+1} \cdot r^{k-1}$$
Work Step by Step
All the terms involve $r$ so replace $1$ with its equivalent $r^0$ to obtain:
$=r^0-r+r^2-r^3+r^4-r^5$
The terms alternate signs so the summation involves either $(-1)^k$ or $(-1)^{k+1}$.
However, since the first term is positive, then the summation must involve $(-1)^{k+1}$ as this gives a positive value when $k=1$.
The summation has 6 terms.
This means that $k$ will vary from $k=1$ to $k=6$.
The exponent of $r$ varies from $0$ to $5$.
This means that when $k=1$, the exponent pf $r$ will be $k-1=0$.
Thus, the exponent of $r$ is $k-1$.
Therefore, the summation notation whose expanded form is given is:
$$\sum_{k=1}^{6} (-1)^{k+1} \cdot r^{k-1}$$