1. **State the problem:** Calculate the sum $$\sum_{r=4}^{n-1} (6r^5 + 2^r + 7)$$ for a general upper limit $n-1$. 2. **Break down the sum:** The sum can be separated into three sums: $$\sum_{r=4}^{n-1} 6r^5 + \sum_{r=4}^{n-1} 2^r + \sum_{r=4}^{n-1} 7$$ 3. **Sum each part:** - For the polynomial term: $$6 \sum_{r=4}^{n-1} r^5$$ - For the geometric term: $$\sum_{r=4}^{n-1} 2^r$$ - For the constant term: $$7 \sum_{r=4}^{n-1} 1 = 7((n-1) - 4 + 1) = 7(n-4)$$ 4. **Formula for sum of powers:** The sum of fifth powers from 1 to $m$ is given by a known formula, but here we need from $r=4$ to $r=n-1$: $$\sum_{r=4}^{n-1} r^5 = \sum_{r=1}^{n-1} r^5 - \sum_{r=1}^{3} r^5$$ 5. **Sum of fifth powers formula:** $$\sum_{r=1}^m r^5 = \left(\frac{m(m+1)}{2}\right)^2 \cdot \frac{2m^2 + 2m - 1}{3}$$ 6. **Apply this:** $$\sum_{r=4}^{n-1} r^5 = \left(\frac{(n-1)n}{2}\right)^2 \cdot \frac{2(n-1)^2 + 2(n-1) - 1}{3} - \left(\frac{3 \cdot 4}{2}\right)^2 \cdot \frac{2 \cdot 3^2 + 2 \cdot 3 - 1}{3}$$ 7. **Sum of geometric series:** $$\sum_{r=4}^{n-1} 2^r = \sum_{r=0}^{n-1} 2^r - \sum_{r=0}^{3} 2^r = (2^{n} - 1) - (2^4 - 1) = 2^{n} - 2^4 = 2^{n} - 16$$ 8. **Combine all parts:** $$\sum_{r=4}^{n-1} (6r^5 + 2^r + 7) = 6 \left[\sum_{r=4}^{n-1} r^5\right] + (2^{n} - 16) + 7(n-4)$$ This expression gives the sum in terms of $n$. **Final answer:** $$\sum_{r=4}^{n-1} (6r^5 + 2^r + 7) = 6 \left[ \left(\frac{(n-1)n}{2}\right)^2 \cdot \frac{2(n-1)^2 + 2(n-1) - 1}{3} - 100 \right] + 2^{n} - 16 + 7(n-4)$$ where $100$ is the value of the sum of fifth powers from 1 to 3 calculated explicitly.