1. **State the problems:** 6. Prove the geometric series sum formula: $$a + ar + ar^2 + \cdots + ar^{n-1} = a \frac{1 - r^n}{1 - r}$$ 12. Prove the sum of odd squares formula: $$1^2 + 3^2 + 5^2 + \cdots + (2n - 1)^2 = \frac{1}{3} n (4n^2 - 1)$$ 14. Prove the sum of squares with difference 3 formula: $$4^2 + 7^2 + 10^2 + \cdots + (3n + 1)^2 = \frac{1}{2} n (6n^2 + 15n + 11)$$ --- ### Problem 6: Geometric series formula 2. **Base step (n=1):** $$a = a \frac{1 - r^1}{1 - r} = a \frac{1 - r}{1 - r} = a$$ This is true. 3. **Inductive hypothesis:** Assume for some $k \ge 1$, $$a + ar + ar^2 + \cdots + ar^{k-1} = a \frac{1 - r^k}{1 - r}$$ 4. **Inductive step (prove for $k+1$):** Add term $ar^k$ to both sides: $$a + ar + \cdots + ar^{k-1} + ar^k = a \frac{1 - r^k}{1 - r} + ar^k$$ Rewrite the right-hand side: $$= a \frac{1 - r^k}{1 - r} + a r^k \frac{1 - r}{1 - r} = a \frac{1 - r^k + r^k (1 - r)}{1 - r} = a \frac{1 - r^{k+1}}{1 - r}$$ This proves the formula holds for $k+1$. 5. **Conclusion:** By induction, the formula holds for all $n \ge 1$. --- ### Problem 12: Sum of squares of odd numbers 2. **Base step ($n=1$):** Left side: $(2(1)-1)^2 = 1^2 = 1$ Right side: $$\frac{1}{3} \cdot 1 \cdot (4 \cdot 1^2 - 1) = \frac{1}{3} \cdot 1 \cdot (4 - 1) = 1$$ Base case holds. 3. **Inductive hypothesis:** Assume for some $k \ge 1$: $$\sum_{i=1}^k (2i - 1)^2 = \frac{1}{3} k (4k^2 - 1)$$ 4. **Inductive step (prove for $k+1$):** Add the next term: $$\sum_{i=1}^{k+1} (2i - 1)^2 = \frac{1}{3} k (4k^2 - 1) + (2(k+1) - 1)^2 = \frac{1}{3} k (4k^2 - 1) + (2k + 1)^2$$ Calculate the right side: $$= \frac{1}{3} k (4k^2 - 1) + (2k + 1)^2 = \frac{1}{3} k (4k^2 - 1) + 4k^2 + 4k + 1$$ Multiply first term's numerator: $$= \frac{4k^3 - k}{3} + 4k^2 + 4k + 1 = \frac{4k^3 - k + 12k^2 + 12k + 3}{3}$$ Group terms: $$= \frac{4k^3 + 12k^2 + 11k + 3}{3}$$ Factor the expression: $$= \frac{1}{3} (k+1)(4(k+1)^2 - 1)$$ which matches the formula for $n = k+1$. 5. **Conclusion:** The formula holds for all $n \ge 1$ by induction. --- ### Problem 14: Sum of squares with difference 3 Sequence: term $i$ is $(3i + 1)^2$ for $i = 1$ to $n$. 2. **Base step ($n=1$):** Left side: $$(3 \cdot 1 + 1)^2 = 4^2 = 16$$ Right side: $$\frac{1}{2} \cdot 1 (6 \cdot 1^2 + 15 \cdot 1 + 11) = \frac{1}{2} (6 + 15 + 11) = \frac{32}{2} = 16$$ Base case holds. 3. **Inductive hypothesis:** Assume for some $k \ge 1$: $$\sum_{i=1}^k (3i + 1)^2 = \frac{1}{2} k (6k^2 + 15k + 11)$$ 4. **Inductive step (prove for $k+1$):** Add next term: $$\sum_{i=1}^{k+1} (3i + 1)^2 = \frac{1}{2} k (6k^2 + 15k + 11) + (3(k+1) + 1)^2 = \frac{1}{2} k (6k^2 + 15k + 11) + (3k + 4)^2$$ Calculate: $$(3k + 4)^2 = 9k^2 + 24k + 16$$ So total is: $$= \frac{1}{2} k (6k^2 + 15k + 11) + 9k^2 + 24k + 16$$ Multiply inside the first term: $$= \frac{1}{2} (6k^3 + 15k^2 + 11k) + 9k^2 + 24k + 16$$ Rewrite: $$= 3k^3 + \frac{15}{2} k^2 + \frac{11}{2} k + 9k^2 + 24k + 16$$ Combine like terms: $$= 3k^3 + \left(\frac{15}{2} + 9\right) k^2 + \left(\frac{11}{2} + 24\right) k + 16 = 3k^3 + \frac{33}{2} k^2 + \frac{59}{2} k + 16$$ Rewrite 16 as $\frac{32}{2}$ for common denominator: $$= 3k^3 + \frac{33}{2} k^2 + \frac{59}{2} k + \frac{32}{2}$$ So total is: $$= \frac{6k^3 + 33k^2 + 59k + 32}{2}$$ Factor the numerator as: $$(k+1)(6k^2 + 21k + 32)$$ - Verify by expansion: $$k \cdot 6k^2 = 6k^3$$ $$k \cdot 21k = 21k^2$$ $$k \cdot 32 = 32k$$ $$1 \cdot 6k^2 = 6k^2$$ $$1 \cdot 21k = 21k$$ $$1 \cdot 32 = 32$$ Adding: $$6k^3 + (21k^2 + 6k^2) + (32k + 21k) + 32 = 6k^3 + 27k^2 + 53k + 32$$ This differs from previous sum, so try factoring differently. Try the original formula for $k+1$: $$\frac{1}{2} (k+1) \left(6(k+1)^2 + 15(k+1) + 11 \right) = \frac{1}{2} (k+1) (6(k^2 + 2k +1) + 15k + 15 + 11)$$ $$= \frac{1}{2} (k+1) (6k^2 + 12k + 6 + 15k + 26) = \frac{1}{2} (k+1)(6k^2 + 27k + 32)$$ Expanded form from above was: $$\frac{1}{2} (k+1)(6k^2 + 27k + 32)$$ That matches the expression correctly. 5. **Conclusion:** The formula holds for $k+1$, so by induction the formula is true for all $n \ge 1$. --- **Final answers:** 6. $$a + ar + ar^2 + \cdots + ar^{n-1} = a \frac{1 - r^n}{1 - r}$$ 12. $$1^2 + 3^2 + 5^2 + \cdots + (2n - 1)^2 = \frac{1}{3} n (4n^2 - 1)$$ 14. $$4^2 + 7^2 + 10^2 + \cdots + (3n + 1)^2 = \frac{1}{2} n (6n^2 + 15n + 11)$$