1. **Problem statement:** Prove by mathematical induction that $$4^2 + 7^2 + 10^2 + \dots + (3n + 1)^2 = \frac{1}{2} n (6n^2 + 15n + 11)$$ for all positive integers $n$. 2. **Base step ($n=1$):** Left-hand side (LHS) = $(3\times1 + 1)^2 = 4^2 = 16$ Right-hand side (RHS) = $\frac{1}{2} \times 1 \times (6 \times 1^2 + 15 \times 1 + 11) = \frac{1}{2} \times (6 + 15 + 11) = \frac{1}{2} \times 32 = 16$ Since LHS = RHS, the base step holds. 3. **Inductive hypothesis:** Assume the formula is true for some $k \geq 1$, that is: $$4^2 + 7^2 + 10^2 + \dots + (3k + 1)^2 = \frac{1}{2} k (6k^2 + 15k + 11)$$ 4. **Inductive step:** We need to prove the formula holds for $k+1$: $$4^2 + 7^2 + 10^2 + \dots + (3k + 1)^2 + (3(k+1) + 1)^2 = \frac{1}{2} (k+1) [6(k+1)^2 + 15(k+1) + 11]$$ Substitute the inductive hypothesis: $$\frac{1}{2} k (6k^2 + 15k + 11) + (3k + 4)^2 = \frac{1}{2} (k+1) (6(k+1)^2 + 15(k+1) + 11)$$ Calculate $(3k + 4)^2 = 9k^2 + 24k + 16$. Left side becomes: $$\frac{1}{2} k (6k^2 + 15k + 11) + 9k^2 + 24k + 16$$ Multiply out first term: $$\frac{1}{2} (6k^3 + 15k^2 + 11k) + 9k^2 + 24k + 16 = 3k^3 + \frac{15}{2} k^2 + \frac{11}{2} k + 9k^2 + 24k + 16$$ Combine like terms: $$3k^3 + \left(\frac{15}{2} + 9\right) k^2 + \left(\frac{11}{2} + 24\right) k + 16 = 3k^3 + \frac{33}{2} k^2 + \frac{59}{2} k + 16$$ Right side: $$\frac{1}{2} (k+1) [6(k+1)^2 + 15(k+1) + 11]$$ Calculate inside bracket: $$6(k^2 + 2k + 1) + 15(k+1) + 11 = 6k^2 + 12k + 6 + 15k + 15 + 11 = 6k^2 + 27k + 32$$ Hence right side: $$\frac{1}{2} (k+1)(6k^2 + 27k + 32) = \frac{1}{2} (6k^3 + 27k^2 + 32k + 6k^2 + 27k + 32) = \frac{1}{2} (6k^3 + 33k^2 + 59k + 32)$$ Simplify: $$3k^3 + \frac{33}{2} k^2 + \frac{59}{2} k + 16$$ The left side equals the right side. 5. **Conclusion:** By the principle of mathematical induction, the formula $$4^2 + 7^2 + 10^2 + \dots + (3n + 1)^2 = \frac{1}{2} n (6n^2 + 15n + 11)$$ holds for all positive integers $n$.