Fill in the blanks with right, left, or midpoint; an interval, and a value of \( n \). \( \sum_{k=1}^{6}(1+k) \cdot 1 \) is a Riemann sum for \( f \) on the interval \( [1, \ldots] \) with \( n= \) Complete the answers below. \( \sum_{k=1}^{6} r(1+k) \cdot 1 \) is a \( \square \) Riemann sum for \( f \) on the interval \( [1, \square \) with \( n= \)

To fill in the blanks, we need to analyze the given Riemann sum: \[ \sum_{k=1}^{6}(1+k) \cdot 1 \] 1. **Identify the interval**: The expression \( (1+k) \) suggests that we are evaluating the function \( f(x) = x \) at points \( x = 1+k \) for \( k = 1, 2, \ldots, 6 \). This means the values of \( x \) will be \( 2, 3, 4, 5, 6, 7 \). Therefore, the interval is from \( 1 \) to \( 7 \). 2. **Determine the type of Riemann sum**: The sum is of the form \( \sum_{k=1}^{n} f(x_k) \cdot \Delta x \), where \( \Delta x = 1 \) and \( x_k = 1+k \). Since we are using the points \( 2, 3, 4, 5, 6, 7 \) (which are the right endpoints of the subintervals), this indicates that it is a right Riemann sum. 3. **Determine the value of \( n \)**: The sum runs from \( k=1 \) to \( k=6 \), which means \( n = 6 \). Now we can fill in the blanks: - The interval is \( [1, 7] \). - The type of Riemann sum is "right". - The value of \( n \) is \( 6 \). Thus, the completed statement is: \[ \sum_{k=1}^{6} r(1+k) \cdot 1 \text{ is a } \text{right} \text{ Riemann sum for } f \text{ on the interval } [1, 7] \text{ with } n= 6. \]