1) Evaluate the following: \( \begin{array}{ll}\text { a) } \sum_{k=1}^{12} 5+2(k-3) & \text { e) } \sum_{k=1}^{12}-12+(r-2)(-2) \\ \text { b) } \sum_{k=5}^{12} 3(2)^{k-5} & \text { d) } \sum_{k=0}^{3} 2(3)(3)^{k}\end{array} \)
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Let's evaluate each of the sums step by step. **a)** \[ \sum_{k=1}^{12} 5 + 2(k-3) = \sum_{k=1}^{12} (5 + 2k - 6) = \sum_{k=1}^{12} (2k - 1). \] This simplifies to: \[ \sum_{k=1}^{12} 2k - \sum_{k=1}^{12} 1 = 2\sum_{k=1}^{12} k - 12 = 2 \left(\frac{12 \cdot 13}{2}\right) - 12 = 156 - 12 = 144. \] **b)** \[ \sum_{k=5}^{12} 3(2)^{k-5} = 3 \sum_{k=5}^{12} (2)^{k-5}. \] Changing the index of summation, let \(j = k - 5\): \[ = 3 \sum_{j=0}^{7} (2)^j = 3(2^8 - 1) = 3(256 - 1) = 3 \cdot 255 = 765. \] **c)** (No part labeled "c" was provided.) **d)** \[ \sum_{k=0}^{3} 2(3)(3)^k = 6 \sum_{k=0}^{3} (3)^k. \] Using the formula for the sum of a geometric series, we find: \[ = 6 \left(\frac{3^{4}-1}{3-1}\right) = 6 \left(\frac{81 - 1}{2}\right) = 6 \cdot 40 = 240. \] **e)** \[ \sum_{k=1}^{12} -12 + (k-2)(-2) = \sum_{k=1}^{12} (-12 - 2k + 4) = \sum_{k=1}^{12} (-2k - 8). \] This simplifies to: \[ -2\sum_{k=1}^{12} k - 8 \cdot 12 = -2 \left(\frac{12 \cdot 13}{2}\right) - 96 = -78 - 96 = -174. \] Summarizing the results of the sums: - a) 144 - b) 765 - d) 240 - e) -174