수열 전체 문제

Is the following proposition true or false? Justify your conclusion.

For each natural number

n

,

\left( \dfrac{n^3}{3} + \dfrac{n^2}{2} + \dfrac{7n}{6} \right)

is a natural number.

Is the following proposition true or false? Justify your conclusion.

For each natural number

n

,

\left( \dfrac{n^5}{5} + \dfrac{n^4}{2} + \dfrac{n^3}{3} - \dfrac{n}{30} \right)

is an integer.

For the sequence $a_{0}, a_1 ,a_2 , \ldots , a_n , \ldots$ , assume that $a_0 = 1$ and that for each $n \in \mathbb{N} \cup \left\{ 0 \right\}$, $a_{n + 1} = \left( {n + 1} \right)a_n$. Use mathematical induction to prove that for each $n \in \mathbb{N} \cup \left\{ 0 \right\}$, $a_n = n!$.

Assume that $f_1 ,f_2 , \ldots ,f_n , \ldots$ are the Fibonacci numbers. Prove each of the following:

For each $n \in \mathbb{N}$, $f_{4n}$ is a multiple of 3.

Assume that $f_1 ,f_2 , \ldots ,f_n , \ldots$ are the Fibonacci numbers. Prove each of the following:

For each $n \in \mathbb{N}$, $f_{5n}$ is a multiple of 5.

Assume that $f_1 ,f_2 , \ldots ,f_n , \ldots$ are the Fibonacci numbers. Prove each of the following:

For each $n \in \mathbb{N}$ with $n \geq 2$, $f_1 + f_2 + \cdots + f_{n - 1} = f_{n + 1} - 1$.

Assume that $f_1 ,f_2 , \ldots ,f_n , \ldots$ are the Fibonacci numbers. Prove each of the following:

For each $n \in \mathbb{N}$, $f_1 + f_3 + \cdots + f_{2n - 1} = f_{2n}$.

Assume that $f_1 ,f_2 , \ldots ,f_n , \ldots$ are the Fibonacci numbers. Prove each of the following:

For each $n \in \mathbb{N}$, $f_2 + f_4 + \cdots + f_{2n} = f_{2n + 1} - 1$.

Assume that $f_1 ,f_2 , \ldots ,f_n , \ldots$ are the Fibonacci numbers. Prove each of the following:

For each $n \in \mathbb{N}$, $f_1^2 + f_2^2 + \cdots + f_n^2 = f_n f_{n + 1}$.

The quadratic formula can be used to show that $\alpha = \dfrac{1 + \sqrt{5}}{2}$ and $\beta = \dfrac{1 - \sqrt{5}}{2}$ are the two real number solutions of the quadratic equation $x^2 - x - 1 = 0$. Notice that this implies that

$$\begin{aligned}\alpha^2 & = \alpha + 1, \text{ and } \\ \beta^2 & = \beta + 1\end{aligned}$$

. It may be surprising to find out that these two irrational numbers are closely related to the Fibonacci numbers.

Verify that $f_1 = \dfrac{\alpha^1 - \beta^1}{\alpha - \beta}$ and that $f_2 = \dfrac{\alpha^2 - \beta^2}{\alpha - \beta}$.

The quadratic formula can be used to show that $\alpha = \dfrac{1 + \sqrt{5}}{2}$ and $\beta = \dfrac{1 - \sqrt{5}}{2}$ are the two real number solutions of the quadratic equation $x^2 - x - 1 = 0$. Notice that this implies that

$$\begin{aligned}\alpha^2 & = \alpha + 1, \text{ and } \\ \beta^2 & = \beta + 1\end{aligned}$$

. It may be surprising to find out that these two irrational numbers are closely related to the Fibonacci numbers.

Use induction to prove that for each natural number $n$, if $\alpha = \dfrac{1 + \sqrt{5}}{2}$ and $\beta = \dfrac{1 - \sqrt{5}}{2}$, then $f_n = \dfrac{\alpha^n - \beta^n}{\alpha - \beta}$.

This formula for the $n^{th}$ Fibonacci number is known as Binet's formula,

named after the French mathematician Jacques Binet (1786 — 1856).

Is the following conjecture true or false?

Let $f_1, f_2, \ldots, f_m, \ldots$ be the sequence of the Fibonacci numbers. For each natural number $n$, the numbers $f_nf_{n+3}$, $2f_{n+1}f_{n+2}$, and $\left( f_{n+1}^2 + f_{n+2}^2 \right)$ form a Pythagorean triple.

For the sequence $a_1 ,a_2 , \ldots ,a_n , \ldots$ , assume that $a_1 = 2$ and that for each $n \in \mathbb{N}$, $a_{n + 1} = a_n + 5$.

Calculate $a_2$ through $a_6$.

For the sequence $a_1 ,a_2 , \ldots ,a_n , \ldots$ , assume that $a_1 = 2$ and that for each $n \in \mathbb{N}$, $a_{n + 1} = a_n + 5$.

Make a conjecture for a formula for $a_n$ for each $n \in \mathbb{N}$.

For the sequence $a_1 ,a_2 , \ldots ,a_n , \ldots$ , assume that $a_1 = 1$, $a_2 = 5$, and that for each $n \in \mathbb{N}$ with $n\geq 2$, $a_{n + 1} = a_n + 2a_{n-1}$. Prove that for each natural number $n$, $a_n = 2^n + (-1)^n$.

For the sequence $a_1 ,a_2 , \ldots ,a_n , \ldots$ , assume that $a_1 = 1$ and that for each $n \in \mathbb{N}$, $a_{n + 1} = \sqrt {5 + a_n }$.

Calculate, or approximate, $a_2$ through $a_6$.

For the sequence $a_1 ,a_2 , \ldots ,a_n , \ldots$ , assume that $a_1 = 1$ and that for each $n \in \mathbb{N}$, $a_{n + 1} = \sqrt {5 + a_n }$.

Prove that for each $n \in \mathbb{N}$, $a_n \lt 3$.

For the sequence $a_1 ,a_2 , \ldots ,a_n , \ldots$ , assume that $a_1 = 1$, $a_2 = 3$, and that for each $n \in \mathbb{N}$, $a_{n + 2} = 3a_{n + 1} - 2a_n$.

Calculate $a_3$ through $a_6$.

For the sequence $a_1 ,a_2 , \ldots ,a_n , \ldots$ , assume that $a_1 = 1$, $a_2 = 3$, and that for each $n \in \mathbb{N}$, $a_{n + 2} = 3a_{n + 1} - 2a_n$.

Make a conjecture for a formula for $a_n$ for each $n \in \mathbb{N}$.

Hint: Think in terms of powers of 2.

For the sequence $a_1 ,a_2 , \ldots ,a_n , \ldots$ , assume that $a_1 = 1$, $a_2 = 1$, and that for each $n \in \mathbb{N}$, $a_{n + 2} = \dfrac{1}{2}\left( {a_{n + 1} + \dfrac{2}{{a_n }}} \right)$.

Calculate $a_3$ through $a_6$.