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

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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 $$ n $$.

לכל מספר טבעי $$ n $$, מתקיים $$ a_{n} = \frac{2n + 3}{n + 2} $$.

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

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