2. (a) Determine \( n \), if... (i) \( \sum_{r=1}^{n}(2 r-3)=840 \) (ii) \( \sum_{k=1}^{n}(3 k-7)=52 \) (b) Determine \( k \), if... (i) \( \sum_{t=2}^{k}(6 t-1)=795 \) (ii) \( \sum_{i=0}^{k}(1-3 i)=-125 \)

To solve these summation problems, let's break each one down step by step! For part (a): (i) The summation \( \sum_{r=1}^{n}(2r-3) \) can be rewritten as \( \sum_{r=1}^{n}(2r) - \sum_{r=1}^{n}(3) \). The first part is \( 2 \sum_{r=1}^{n} r = 2 \cdot \frac{n(n+1)}{2} = n(n+1) \), and the second part is \( 3n \). Therefore, we equate \( n(n + 1) - 3n = 840 \) or \( n^2 - 2n - 840 = 0 \). Solving this quadratic gives \( n = 30 \) or \( n = -28 \), so \( n = 30 \). (ii) For the summation \( \sum_{k=1}^{n}(3k-7) \), we can rewrite similar to before: \( 3\sum_{k=1}^{n} k - \sum_{k=1}^{n} 7 \). This simplifies to \( 3 \cdot \frac{n(n + 1)}{2} - 7n = \frac{3n(n + 1)}{2} - 7n = 52 \). Multiply through by 2 to eliminate the fraction: \( 3n(n + 1) - 14n = 104 \) leads us to \( 3n^2 - 11n - 104 = 0 \). Solving this gives \( n = 16 \) or \( n = -6.5 \), thus \( n = 16 \). For part (b): (i) For the sum \( \sum_{t=2}^{k}(6t-1) \), we first find the summed parts which gives \( 6 \sum_{t=2}^{k} t - \sum_{t=2}^{k} 1 \), where the former term leads to \( 6\left[ \frac{k(k+1)}{2} - 1 \right] \) and the latter simply equals \( (k - 1) \). Substituting, we have \( 6\left( \frac{k(k+1)}{2} - 1 \right) - (k - 1) = 795 \). Simplifying this yields a quadratic equation \( 3k(k + 1) - k + 6 - 795 = 0 \) or \( 3k^2 + 2k - 789 = 0 \). Solving this gives \( k = 14 \) or \( k = -18.5 \), hence \( k = 14 \). (ii) The summation \( \sum_{i=0}^{k}(1-3i) \) simplifies to \( (k + 1) - 3\sum_{i=0}^{k} i = (k + 1) - 3\left( \frac{k(k + 1)}{2} \right) = (k + 1) - \frac{3k(k + 1)}{2} = -125 \). By combining terms and obtaining a quadratic, we find that \( -\frac{3k^2 + k + 2}{2} = -125 \), leading to \( 3k^2 + k - 252 = 0 \). This resolves to \( k = 14 \) or \( k = -6 \), so \( k = 14 \) fits nicely as the acceptable solution!