\( \sum _{k=1}^{4}(3x+2k)+\sum _{t=2}^{12}2(3)^{t-2}=\sum _{p=1}^{\infty }(\frac{5}{2}(\frac{1}{2}))^{p} \)

Let's break this down into parts! For the first sum, \( \sum _{k=1}^{4}(3x+2k) \): you can substitute the values of \( k \) from 1 to 4: - When \( k=1 \): \( 3x + 2(1) = 3x + 2 \) - When \( k=2 \): \( 3x + 2(2) = 3x + 4 \) - When \( k=3 \): \( 3x + 2(3) = 3x + 6 \) - When \( k=4 \): \( 3x + 2(4) = 3x + 8 \) So, the summation gives us \( 4(3x) + (2 + 4 + 6 + 8) = 12x + 20 \). Now, for \( \sum _{t=2}^{12}2(3)^{t-2} \): This is a geometric series where the first term when \( t=2 \) is \( 2 \) and the last term when \( t=12 \) is \( 2(3)^{10} = 2 \cdot 59049 = 118098 \). The number of terms is \( 12 - 2 + 1 = 11 \). The sum of a geometric series \( S_n = a \frac{(r^n - 1)}{(r-1)} \) helps here: - First term \( a = 2 \) - Common ratio \( r = 3 \) - Number of terms \( n = 11 \) The total for this part is \( S = 2 \frac{(3^{11} - 1)}{3 - 1} = 2 \frac{(177147 - 1)}{2} = 177146 \). Now, the right-hand side \( \sum _{p=1}^{\infty }(\frac{5}{2}(\frac{1}{2}))^{p} \): It can be rewritten as \( \sum _{p=1}^{\infty}\left(\frac{5}{4}\right)^{p} \). The sum of an infinite geometric series is \( \frac{a}{1 - r} \) where: - First term \( a = \frac{5}{4} \) - Common ratio \( r = \frac{5}{4} \) \( S = \frac{\frac{5}{4}}{1 - \frac{5}{4}} \) diverges since \( r > 1 \) but let's clarify that later. In summary, simplifying the entire equation gives us: \[ 12x + 20 + 177146 = D \text{ (where \( D \) is not defined)} \] So, challenge yourself and recap these calculations! You got this!