24 (1152). Sequences $\left(x_{n}\right)$ and ( $y_{n}$ ) are defined by the formulas $x_{n}=2 n-1$ and $y_{n}=n^{2}$. If all common terms of these sequences are written in ascending order, the resulting sequence is $\left(c_{n}\right)$. Write the formula for the $n$-th term of the sequence $\left(c_{n}\right)$.
Solution. Let's write the sequences $\left(x_{n}\right),\left(y_{n}\right)$ and $\left(c_{n}\right)$:

$$
\begin{aligned}
& \left(x_{n}\right) \quad 1,3,5,7,9, \ldots ; \\
& \left(y_{n}\right) \quad 1,4,9,16,25,36,49, \ldots ; \\
& \left(c_{n}\right) \quad 1,9,25,49, \ldots
\end{aligned}
$$

Thus, the terms of the sequence $\left(c_{n}\right)$ are the terms of the sequence $y_{n}$ with odd indices, i.e., $c_{n}=(2 n-1)^{2}$.

Remark. The formula for $c_{n}$ can also be obtained using other reasoning. We have:

$$
\begin{aligned}
& c_{1}=1=1^{2} \\
& c_{2}=9=3^{2}=(n+1)^{2} \\
& c_{3}=25=5^{2}=(n+2)^{2} \\
& c_{4}=49=7^{2}=(n+3)^{2}
\end{aligned}
$$

Thus, $c_{n}=(n+(n-1))^{2}=(2 n-1)^{2}$. (It is advisable to prove the obtained formula for $c_{n}$ using mathematical induction.)

Answer: $(2 n-1)^{2}$.

口 25(1153). For which values of $n$ do the terms of the sequence defined by the formula $x_{n}=(n+4)(n-5)$ satisfy the condition $-18 \leqslant x_{n} \leqslant 360$?

Solution. Since $x_{n}=n^{2}-n-20$ and $-18 \leqslant x_{n} \leqslant 360$, the problem reduces to solving the system of inequalities:

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-170.jpg?height=123&width=908&top_left_y=1572&top_left_x=580)

Solving the first inequality in the system, we find $-19 \leqslant n \leqslant 20$, and the values of $n$ that satisfy the condition of the problem $(n \in \boldsymbol{N})$ are $n \in[1,20](*)$.

The solution to the second inequality are the values of $n$ that satisfy the inequality $n \leqslant-1$ or the inequality $n \geqslant 2$, of which only $n \in[2 ;+\infty)(* *)$ satisfy the condition of the problem $(n \in \boldsymbol{N})$.

Considering the solutions (*) and (**), we find the answer to the problem.

Answer: $2 \leqslant n \leqslant 20, n \in \boldsymbol{N}$.

Remark. It is advisable to perform a check.

For $n=2, x_{n}=-18$; for $n=20, x_{n}=360$. Since $x_{n}=n^{2}-n-20$ is a continuous function, for $2 \leqslant n \leqslant 20$ we have $-18 \leqslant x_{n} \leqslant 360$.

$\triangle 26$ (1154). Find the sum of the first $n$ terms of the sequence $\left(x_{n}\right)$, if $x_{n}=\frac{1}{(2 n-1)(2 n+1)}$.

Solution. $x_{1}=\frac{1}{3}, x_{2}=\frac{1}{3 \cdot 5}=\frac{1}{15}, x_{3}=\frac{1}{5 \cdot 7}=\frac{1}{35}$, $x_{4}=\frac{1}{7 \cdot 9}, x_{5}=\frac{1}{9 \cdot 11}$.

$S_{n}=\frac{1}{3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\frac{1}{7 \cdot 9}+\frac{1}{9 \cdot 11}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=$ $=\frac{1}{3}+\frac{1}{2} \cdot\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2} \cdot\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2} \cdot\left(\frac{1}{9}-\frac{1}{11}\right)+\ldots+$ $+\frac{1}{2}\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)=\frac{1}{3}+\frac{1}{6}-\frac{1}{2(2 n+1)}=\frac{1}{2}-\frac{1}{2(2 n+1)}=$

$$
=\frac{2 n+1-1}{2(2 n+1)}=\frac{2 n}{2(2 n+1)}=\frac{n}{2 n+1}
$$

Answer: $\frac{n}{2 n+1}$.

Remark 1. If students have difficulty solving this problem, it is advisable to suggest the following auxiliary problem: "Find the value of the expression

$$
\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{19 \cdot 20} »
$$

Solution.

$$
\begin{gathered}
\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{19 \cdot 20}= \\
=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\ldots+\frac{1}{19}-\frac{1}{20}= \\
=1-\frac{1}{20}=\frac{19}{20}
\end{gathered}
$$

Remark 2. It is advisable to perform a check by calculating several values of $S_{n}$ using the formula and by directly summing the first $n$ terms. For example, for $n=1$ the formula gives $S_{1}=\frac{1}{3}$. By the condition $x_{1}=\frac{1}{3}, S_{1}=x_{1}$.

For $n=2, x_{1}+x_{2}=\frac{1}{3}+\frac{1}{15}=\frac{2}{5}$. By the formula $S_{2}=\frac{2}{5}$.

For $n=3, x_{1}+x_{2}+x_{3}=\frac{2}{5}+\frac{1}{35}=\frac{3}{7}$. By the formula $S_{3}=\frac{3}{7}$.

27(1155). In the sequence $\left(x_{n}\right)$, each term with an odd index is equal to $2 a$, and each term with an even index is equal to $2 b$. Write the formula for the $n$-th term of this sequence.

Solution. The given sequence has the form $2 a, 2 b, 2 a, 2 b, \ldots$. The formula for the $n$-th term of this sequence can be given in several ways.

First method. We show that the sequence $\left(x_{n}\right)$ can be defined by the formula $x_{n}=c+(-1)^{n} d$, where $c$ and $d$ are some numbers. Then, by the condition, we have:

$$
\left\{\begin{array}{l}
c+d=2 b \\
c-d=2 a
\end{array}\right.
$$

from which $c=a+b, d=b-a$. Therefore,

$$
x_{n}=(a+b)+(-1)^{n} \cdot(b-a), n \in \boldsymbol{N}
$$

Remark. After obtaining the formula $x_{n}=(a+b)+(-1)^{n}(b-a)$, it is advisable to suggest that students write it in another way using information from trigonometry. We get:

a) $x_{n}=(b-a) \sin \left(\frac{\pi}{2}(2 n+1)\right)+(a+b), n \in \boldsymbol{N}$;

b) $x_{n}=(b-a) \cos \pi n+(a+b), n \in \boldsymbol{N}$.

Second method. If students are familiar with the concept of the fractional part of a number, then the formula for the $n$-th term of the given sequence can be written, for example, as follows:

$$
x_{2 n+1}=4 a\left\{\frac{n}{2}\right\}+4 b\left\{\frac{n+1}{2}\right\}
$$

where $\left\{\frac{n}{2}\right\}$ is the fractional part of the number $\frac{n}{2}$.

$\left\{\frac{n+1}{2}\right\}$ is the fractional part of the number $\frac{n+1}{2}$.

Indeed, we have:

$$
\begin{gathered}
x_{1}=4 a\left\{\frac{1}{2}\right\}+4 b\{1\}=4 a \cdot \frac{1}{2}+4 a \cdot 0=2 a \\
x_{2}=4 a\{1\}+4 b\left\{\frac{3}{2}\right\}=4 a \cdot 0+4 b \cdot \frac{1}{2}=2 b \\
x_{3}=4 a