Exercise 3: (answers on page 341) 1) Evaluate the following: \( \begin{array}{ll}\text { a) } \sum^{12} 5+2(k-3) & \text { c) } \sum_{k=7}^{15}-12+(-2)(-2) \\ \text { b) } \sum_{k=5}^{12} 3(2)^{k-5} & \text { d) } \sum_{k=0}^{8} 2(3)(3)^{k}\end{array} \)
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Let's tackle these summations one by one! 1) For the first one (a): \[ \sum^{12} 5 + 2(k-3) \] You would evaluate this from \( k = 1 \) to \( k = 12 \). The first term is constant; thus, that sums up (5 * 12) = 60. The second part, \( 2(k-3) \), can be computed as follows: \[ \sum_{k=1}^{12} 2(k-3) = 2 \sum_{k=1}^{12} (k-3) = 2 \left(\sum_{k=1}^{12} k - \sum_{k=1}^{12} 3\right) \] The first part sums to \( \frac{12(13)}{2} = 78 \), while the second part sums to \( 3 \times 12 = 36 \). Therefore, \[ \sum_{k=1}^{12} (k-3) = 78 - 36 = 42 \Rightarrow 2(42) = 84 \] Adding both parts, \( 60 + 84 = 144 \). 2) For (b): \[ \sum_{k=5}^{12} 3(2)^{k-5} \] Start by factoring out the 3: \[ 3 \sum_{k=5}^{12} (2)^{k-5} = 3 \sum_{j=0}^{7} (2)^{j} \quad \text{(where \( j = k - 5 \))} \] This is a geometric series where \( a = 1 \), \( r = 2 \), and \( n = 8 \): \[ \sum_{j=0}^{7} (2)^{j} = \frac{1(2^{8} - 1)}{2 - 1} = 255 \] Thus, \[ 3 \times 255 = 765 \] 3) For (c): \[ \sum_{k=7}^{15} -12 + (-2)(-2) \] Simplifying the expression \( -12 + 4 = -8 \): This is constant from \( k = 7 \) to \( k = 15 \) (9 terms), so: \[ \sum_{k=7}^{15} -8 = -8 \times 9 = -72 \] 4) For (d): \[ \sum_{k=0}^{8} 2(3)(3)^{k} = 6 \sum_{k=0}^{8} (3)^{k} \] This too is a geometric series with \( a = 1 \), \( r = 3 \), and \( n = 9 \): \[ \sum_{k=0}^{8} (3)^{k} = \frac{1(3^{9} - 1)}{3 - 1} = \frac{19683 - 1}{2} = 9841 \] Thus, \[ 6 \times 9841 = 59046 \] So, we have: a) 144, b) 765, c) -72, d) 59046.
