\( a_{n+1}=a_{n}+\frac{1}{(n+2)(n+3)} \quad \) 5תונה סדרה המוגדרת כך : \( a_{1}=\frac{5}{3} \) ולכל טבעי, הוכיחו באינדוקציה כי לכל \( n \) טבעי מתקיים: \( a_{n}=\frac{2 n+3}{n+2} \)

Alright, I need to prove by induction that for the sequence defined by \( a_{1} = \frac{5}{3} \) and \( a_{n+1} = a_{n} + \frac{1}{(n+2)(n+3)} \) for all natural numbers \( n \), the formula \( a_{n} = \frac{2n + 3}{n + 2} \) holds true for every natural number \( n \). First, let's understand the sequence and the formula. The sequence starts with \( a_1 = \frac{5}{3} \), and each subsequent term is obtained by adding \( \frac{1}{(n+2)(n+3)} \) to the previous term. The formula to be proven suggests that each term \( a_n \) can be expressed as \( \frac{2n + 3}{n + 2} \). To prove this by induction, I need to follow two main steps: the base case and the inductive step. **Base Case:** Let's verify the formula for \( n = 1 \). Given \( a_1 = \frac{5}{3} \), and according to the formula, \( a_1 = \frac{2(1) + 3}{1 + 2} = \frac{5}{3} \). The formula holds for \( n = 1 \). **Inductive Step:** Assume that the formula holds for some arbitrary natural number \( k \), i.e., \( a_k = \frac{2k + 3}{k + 2} \). Now, I need to show that the formula also holds for \( k + 1 \), i.e., \( a_{k+1} = \frac{2(k + 1) + 3}{(k + 1) + 2} = \frac{2k + 5}{k + 3} \). Starting with the recursive definition: \[ a_{k+1} = a_k + \frac{1}{(k + 2)(k + 3)} \] Substituting the inductive hypothesis \( a_k = \frac{2k + 3}{k + 2} \): \[ a_{k+1} = \frac{2k + 3}{k + 2} + \frac{1}{(k + 2)(k + 3)} \] To combine these fractions, find a common denominator: \[ a_{k+1} = \frac{(2k + 3)(k + 3) + 1}{(k + 2)(k + 3)} \] Expanding the numerator: \[ (2k + 3)(k + 3) = 2k^2 + 6k + 3k + 9 = 2k^2 + 9k + 9 \] So, \[ a_{k+1} = \frac{2k^2 + 9k + 9 + 1}{(k + 2)(k + 3)} = \frac{2k^2 + 9k + 10}{(k + 2)(k + 3)} \] Now, let's simplify the numerator: \[ 2k^2 + 9k + 10 = 2k^2 + 4k + 5k + 10 = 2k(k + 2) + 5(k + 2) = (2k + 5)(k + 2) \] Therefore, \[ a_{k+1} = \frac{(2k + 5)(k + 2)}{(k + 2)(k + 3)} = \frac{2k + 5}{k + 3} \] Which matches the formula \( a_{k+1} = \frac{2(k + 1) + 3}{(k + 1) + 2} = \frac{2k + 5}{k + 3} \). **Conclusion:** By the principle of mathematical induction, since the formula holds for \( n = 1 \) and assuming it holds for \( n = k \) leads to it holding for \( n = k + 1 \), the formula \( a_n = \frac{2n + 3}{n + 2} \) is valid for all natural numbers \( n \).