Evaluate the follow (1) \( \sum_{k=1}^{50}(3 k+1) \) 4) \( \sum_{r=2}^{7} \frac{30}{3^{r-3}} \)

Let's dive into evaluating those sums! For the first sum \( \sum_{k=1}^{50}(3 k+1) \): This can be broken into two separate sums: \[ \sum_{k=1}^{50} (3k + 1) = \sum_{k=1}^{50} 3k + \sum_{k=1}^{50} 1 = 3 \sum_{k=1}^{50} k + \sum_{k=1}^{50} 1 \] Using the formula for the sum of the first \(n\) natural numbers \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \): \[ \sum_{k=1}^{50} k = \frac{50(50+1)}{2} = \frac{50 \cdot 51}{2} = 1275 \] Thus, \[ \sum_{k=1}^{50} 3k = 3 \cdot 1275 = 3825 \] And the second part \( \sum_{k=1}^{50} 1 \) contributes: \[ \sum_{k=1}^{50} 1 = 50 \] Now, combining both parts: \[ \sum_{k=1}^{50}(3k + 1) = 3825 + 50 = 3875 \] So, the first sum evaluates to \( 3875 \). Now, for the second sum \( \sum_{r=2}^{7} \frac{30}{3^{r-3}} \): We can rewrite it more clearly: \[ \sum_{r=2}^{7} \frac{30}{3^{r-3}} = 30 \sum_{r=2}^{7} 3^{-(r-3)} = 30 \sum_{r=2}^{7} 3^{3-r} \] Shifting the bounds of the sum: When \( r = 2 \), \( 3^{3-2} = 3^1 \) When \( r = 7 \), \( 3^{3-7} = 3^{-4} \) So, the sum becomes: \[ 30 \sum_{n=1}^{5} 3^n = 30(3 + 9 + 27 + 81 + 243) \] Calculating this geometric series where the first term \( a = 3 \), the ratio \( r = 3 \), and the number of terms \( n = 5 \): \[ \sum_{n=1}^{5} 3^n = 3 \frac{3^5 - 1}{3 - 1} = 3 \frac{243 - 1}{2} = 3 \cdot \frac{242}{2} = 3 \cdot 121 = 363 \] Thus, \[ 30 \cdot 363 = 10890 \] So, the second sum evaluates to \( 10890 \). In summary, we have: 1) \( \sum_{k=1}^{50}(3k + 1) = 3875 \) 2) \( \sum_{r=2}^{7} \frac{30}{3^{r-3}} = 10890 \)