1. **Problem A: Closed form for the sequence** The sequence is given by $a_1 = 1$ and $a_{n+1} = a_n + n + 1$. 2. To find a closed form, rewrite the recurrence: $$a_{n+1} - a_n = n + 1$$ This indicates that the difference between consecutive terms grows linearly. 3. Sum both sides from $n=1$ to $n-1$: $$a_n - a_1 = \sum_{k=1}^{n-1} (k+1) = \sum_{k=1}^{n-1} k + \sum_{k=1}^{n-1} 1 = \frac{(n-1)n}{2} + (n-1)$$ 4. Substitute $a_1=1$: $$a_n = 1 + \frac{(n-1)n}{2} + (n-1) = 1 + \frac{n^2 - n}{2} + n - 1 = \frac{n^2 - n}{2} + n$$ 5. Simplify: $$a_n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}$$ 6. Calculate $a_{100}$: $$a_{100} = \frac{100 \times 101}{2} = 50 \times 101 = 5050$$ --- 7. **Problem B: Showing no real solution for $\sqrt{3+x} + \sqrt{7 - x} = 5$** 8. The domain restrictions are: $$3 + x \geq 0 \implies x \geq -3$$ $$7 - x \geq 0 \implies x \leq 7$$ So $x \in [-3,7]$. 9. Apply Cauchy-Schwarz or analyze maximum of the left side: Notice that $\sqrt{3 + x}$ is increasing and $\sqrt{7 - x}$ is decreasing in $x$. 10. The maximum sum occurs somewhere inside the domain. Let's check endpoints: - At $x = -3$: $\sqrt{0} + \sqrt{10} = 0 + \sqrt{10} \approx 3.162 < 5$ - At $x = 7$: $\sqrt{10} + \sqrt{0} = 3.162 < 5$ 11. Try $x = 2$: $$\sqrt{3 + 2} + \sqrt{7 - 2} = \sqrt{5} + \sqrt{5} = 2\sqrt{5} \approx 4.472 < 5$$ 12. Since the function is continuous and maximum of sum of square roots with $x$ inside the domain never reaches 5, no $x \in \mathbb{R}$ satisfies the equation. --- 13. **Problem C: Evaluate the expression** The expression: $$\sum_{k=1}^3 \frac{\sin\left(\frac{k\pi}{12}\right) + \sin\left(\frac{(6-k)\pi}{12}\right)}{\cos\left(\frac{k\pi}{12}\right) + \cos\left(\frac{(6-k)\pi}{12}\right)} - \prod_{m=1}^2 \frac{\tan\left(\frac{\pi}{4} + \frac{m\pi}{12}\right) - \tan\left(\frac{\pi}{4} - \frac{m\pi}{12}\right)}{1 + \tan\left(\frac{\pi}{4} + \frac{m\pi}{12}\right) \tan\left(\frac{\pi}{4} - \frac{m\pi}{12}\right)}$$ 14. Use sum-to-product identities for numerator and denominator in sums: $$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ $$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ 15. For the summation term inside numerator and denominator with $A = \frac{k\pi}{12}$ and $B = \frac{(6-k)\pi}{12}$: - $\frac{A+B}{2} = \frac{\frac{k\pi}{12} + \frac{(6-k)\pi}{12}}{2} = \frac{6\pi/12}{2} = \frac{\pi}{4}$ - $\frac{A-B}{2} = \frac{\frac{k\pi}{12} - \frac{(6-k)\pi}{12}}{2} = \frac{(2k - 6)\pi/12}{2} = \frac{(k-3)\pi}{12}$ Thus, $$\frac{\sin A + \sin B}{\cos A + \cos B} = \frac{2 \sin(\pi/4) \cos((k-3)\pi/12)}{2 \cos(\pi/4) \cos((k-3)\pi/12)} = \frac{\sin(\pi/4)}{\cos(\pi/4)} = \tan(\pi/4) = 1$$ 16. Each summand is 1 for $k=1,2,3$, so the sum is: $$\sum_{k=1}^3 1 = 3$$ 17. For the product term, recall tangent subtraction formula: $$\frac{\tan X - \tan Y}{1 + \tan X \tan Y} = \tan(X - Y)$$ 18. With $X = \frac{\pi}{4} + \frac{m\pi}{12}$ and $Y = \frac{\pi}{4} - \frac{m\pi}{12}$, we have: $$X - Y = \pi/4 + m\pi/12 - (\pi/4 - m\pi/12) = 2m\pi/12 = \frac{m\pi}{6}$$ 19. So each factor is: $$\tan\left(\frac{m\pi}{6}\right)$$ 20. The product becomes: $$\prod_{m=1}^2 \tan\left(\frac{m\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right) \times \tan\left(\frac{\pi}{3}\right)$$ Recall: $$\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}, \quad \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$$ 21. Therefore: $$\prod = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1$$ 22. Final value of the entire expression is: $$3 - 1 = 2$$ --- **Final answers:** - Problem A: $a_n = \frac{n(n+1)}{2}$ and $a_{100} = 5050$ - Problem B: No real $x$ solves $\sqrt{3+x} + \sqrt{7-x} = 5$ - Problem C: Value of the expression is $2$