Question:

For part (A), $A = \frac{n(n+1)(2n+1)}{6}$ is the sum of squares formula. Part (B) simplifies $B = \sum_{m=1}^{n} \frac{m(m+1)}{2} \cdot \frac{n(n+1)}{2}$ by factoring out constants and applying summation formulas to get $\frac{n(n+1)(n+2)}{6} - \frac{n(n+1)}{4} = \frac{n(n+1)(2n+1)}{12}$. Part (C) uses the geometric series formula with $|x| 0$ and shows $2x^3 - \frac{3}{x^2} = -(2t^3 + \frac{3}{t^2}) \leq -5$ using calculus, leading to $2x^3 + \frac{3}{t^2} \geq 5$. Part (C) applies AM-GM to $\frac{x^3 + x^3 + \cdots}{8}$ to establish $(8-x^3)(x^3) \leq 3(5x^3)^{5/3}$. Part (D) uses AM-GM on the product of two groups: $(x+x+\cdots+x) \cdot (y+y+\cdots+y) \geq (7x^3y^3)^{1/12}$, giving $12a^{1/12} \geq 12$, so $a \geq 1$.