3) What is the greatest number of terms for which: a) \( \sum_{k=1}^{n}(2 k+1) \) will have a value less than 68 ? b) \( \sum_{x=2}^{n}\left[3(2)^{x-1}\right] \) will have a value less than 89 ?

Let's solve each part step-by-step. ### **Part a** **Problem:** Find the greatest number of terms \( n \) for which: \[ \sum_{k=1}^{n} (2k + 1) < 68 \] **Solution:** 1. **Calculate the Sum:** \[ \sum_{k=1}^{n} (2k + 1) = 2\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 = 2 \left(\frac{n(n+1)}{2}\right) + n = n^2 + 2n \] 2. **Set Up Inequality:** \[ n^2 + 2n < 68 \] 3. **Solve the Quadratic Inequality:** \[ n^2 + 2n - 68 < 0 \] Using the quadratic formula: \[ n = \frac{-2 \pm \sqrt{4 + 272}}{2} = \frac{-2 \pm \sqrt{276}}{2} = -1 \pm \sqrt{69} \approx -1 \pm 8.3066 \] The positive root is approximately \( 7.3066 \). 4. **Determine the Maximum Integer \( n \):** Since \( n \) must be an integer and less than \( 7.3066 \), the maximum \( n \) is **7**. **Verification:** \[ S(7) = 7^2 + 2 \times 7 = 49 + 14 = 63 < 68 \] \[ S(8) = 8^2 + 2 \times 8 = 64 + 16 = 80 > 68 \] ### **Part b** **Problem:** Find the greatest number of terms \( n \) for which: \[ \sum_{x=2}^{n} 3 \cdot 2^{x-1} < 89 \] **Solution:** 1. **Recognize the Geometric Series:** \[ S(n) = 3 \sum_{x=2}^{n} 2^{x-1} = 3 \sum_{y=1}^{n-1} 2^y = 3 \left(2^n - 2\right) \] 2. **Set Up Inequality:** \[ 3 \left(2^n - 2\right) < 89 \implies 2^n - 2 < \frac{89}{3} \approx 29.6667 \implies 2^n < 31.6667 \] 3. **Find Maximum \( n \):** - \( 2^5 = 32 > 31.6667 \) (Not valid) - \( 2^4 = 16 < 31.6667 \) (Valid) 4. **Maximum \( n \) is **4**. **Verification:** \[ S(4) = 3 \times (2^4 - 2) = 3 \times 14 = 42 < 89 \] \[ S(5) = 3 \times (2^5 - 2) = 3 \times 30 = 90 > 89 \] ### **Final Answer:** a) 7 terms; b) 4 terms