1. **Problem Statement:** Show that $$\sum_{r=15}^n (6r^2 + 2) = n^3 + (n+1)^3 - k$$ where $k$ is a constant. 2. **Method of Differences:** The method of differences involves expressing the summand as a difference of two terms such that when summed, most terms cancel out, leaving only a few terms from the start and end of the sum. 3. **Rewrite the summand:** We want to find a function $f(r)$ such that $$f(r) - f(r-1) = 6r^2 + 2$$ 4. **Guessing $f(r)$:** Try a cubic polynomial $f(r) = ar^3 + br^2 + cr + d$. Then, $$f(r) - f(r-1) = a(r^3 - (r-1)^3) + b(r^2 - (r-1)^2) + c(r - (r-1))$$ Calculate each difference: $$r^3 - (r-1)^3 = 3r^2 - 3r + 1$$ $$r^2 - (r-1)^2 = 2r - 1$$ $$r - (r-1) = 1$$ So, $$f(r) - f(r-1) = a(3r^2 - 3r + 1) + b(2r - 1) + c$$ 5. **Equate coefficients:** We want $$6r^2 + 2 = 3a r^2 - 3a r + a + 2b r - b + c$$ Group terms: $$6r^2 + 0r + 2 = (3a) r^2 + (-3a + 2b) r + (a - b + c)$$ Equate coefficients: - For $r^2$: $3a = 6 \Rightarrow a = 2$ - For $r$: $-3a + 2b = 0 \Rightarrow -6 + 2b = 0 \Rightarrow 2b = 6 \Rightarrow b = 3$ - Constant term: $a - b + c = 2 \Rightarrow 2 - 3 + c = 2 \Rightarrow c = 3$ 6. **So,** $$f(r) = 2r^3 + 3r^2 + 3r + d$$ 7. **Sum from $r=15$ to $n$:** By telescoping, $$\sum_{r=15}^n (6r^2 + 2) = \sum_{r=15}^n [f(r) - f(r-1)] = f(n) - f(14)$$ 8. **Calculate $f(n)$ and $f(14)$:** $$f(n) = 2n^3 + 3n^2 + 3n + d$$ $$f(14) = 2(14)^3 + 3(14)^2 + 3(14) + d$$ 9. **Rewrite $f(n)$:** Note that $$n^3 + (n+1)^3 = n^3 + (n^3 + 3n^2 + 3n + 1) = 2n^3 + 3n^2 + 3n + 1$$ So, $$f(n) = n^3 + (n+1)^3 + (d - 1)$$ 10. **Therefore,** $$\sum_{r=15}^n (6r^2 + 2) = f(n) - f(14) = [n^3 + (n+1)^3 + (d - 1)] - [f(14)]$$ Let $$k = f(14) - (d - 1)$$ Then, $$\sum_{r=15}^n (6r^2 + 2) = n^3 + (n+1)^3 - k$$ 11. **Summary:** We have shown using the method of differences that the sum can be expressed as the difference of $f(n)$ and $f(14)$, which simplifies to the given form with a constant $k$. --- **Final answer:** $$\sum_{r=15}^n (6r^2 + 2) = n^3 + (n+1)^3 - k$$ where $$k = f(14) - (d - 1)$$ with $f(r) = 2r^3 + 3r^2 + 3r + d$ and $d$ arbitrary constant (cancels out).