## Problem 2.

Determine the strictly positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$, with $a_{1}=1$, for which the following equality holds:

$\frac{3}{a_{1} \cdot a_{2}}+\frac{5}{a_{2} \cdot a_{3}}+\frac{7}{a_{3} \cdot a_{4}}+\ldots+\frac{2 n+1}{a_{n} \cdot a_{n+1}}=1-\frac{1}{a_{n+1}},(\forall) n \in \mathbb{N}^{*}$
## Solution and grading:

$n=1 \Rightarrow \frac{3}{a_{2}}=1-\frac{1}{a_{2}} \Rightarrow a_{2}=4$.................................................................................

$n=2 \Rightarrow \frac{3}{4}+\frac{5}{4 a_{3}}=1-\frac{1}{a_{3}} \Rightarrow a_{3}=9$

![](https://cdn.mathpix.com/cropped/2024_06_07_55fac7dbc24b8ba3d98dg-2.jpg?height=62&width=1499&top_left_y=1965&top_left_x=336)

From the hypothesis, we have $1-\frac{1}{a_{k}}+\frac{2 k+1}{a_{k} \cdot a_{k+1}}=1-\frac{1}{a_{k+1}} \Rightarrow \frac{2 k+1}{a_{k} \cdot a_{k+1}}+\frac{1}{a_{k+1}}=\frac{1}{a_{k}}$. .................1p

![](https://cdn.mathpix.com/cropped/2024_06_07_55fac7dbc24b8ba3d98dg-2.jpg?height=60&width=1502&top_left_y=2106&top_left_x=337)

In conclusion, $a_{n}=n^{2}$, for any non-zero natural number $n$. ..........................................1p