수열 전체 문제

이항정리를 이용하여 ${\left(\frac{3}{2}x-\frac{1}{2}y\right)}^{5}$ 을 전개하세요.

이항식을 완전히 전개하지 않고 ${\left({x}^{2}-\frac{1}{2}\right)}^{13}$ 의 일곱째 항을 구하세요.

Calculate $1 + 2 + 3 + \cdots + n$ and $\dfrac{n(n + 1)}{2}$ for several natural numbers $n$. What do you observe?

Explain the following.

Can a finite, nonempty set be inductive?

Explain the following.

Is the empty set inductive?

Complete the following.

Calculate $1 + 3 + 5 + \cdots + \left( {2n - 1} \right)$ for several natural numbers $n$.

Complete the following.

Calculate the value of $5^n - 2^n$ for $n = 1$, $n = 2$, $n = 3$, $n = 4$, $n = 5$, and $n = 6$.

Formulate a conjecture (with an appropriate quantifier) that can be used as an answer to each of the following questions.

For which natural numbers $n$ is $3^n$ greater than $1 + 2^n$?

Formulate a conjecture (with an appropriate quantifier) that can be used as an answer to each of the following questions.

For which natural numbers $n$ is $2^n$ greater than $\left(n + 1 \right)^2$?

Formulate a conjecture (with an appropriate quantifier) that can be used as an answer to each of the following questions.

For which natural numbers $n$ is $\left( 1 + \dfrac{1}{n} \right)^n$ greater than 2.5?

Consider the following question:

For which natural numbers

n

do there exist nonnegative integers

x

and

y

such that

n = 3x + 5y

?

To help answer this question, we will let $\mathbb{Z}^* = \left\{ {x \in \mathbb{Z} \mid x \geq 0} \right\}$, and let $P \left( n \right)$ be There exist $x,y \in \mathbb{Z}^*$ such that $n = 3x + 5y$. Notice that $P(1)$ is false since if both $x$ and $y$ are zero, then $3x + 5y = 0$ and if either $x > 0$ or $y > 0$, then $3x + 5y \geq 3$. Also notice that $P(6)$ is true since $6 = 3 \cdot 2 + 5 \cdot 0$ and $P(8)$ is true since $8 = 3 \cdot 1 + 5 \cdot 1$.

Explain why $P(2)$, $P(4)$, and $P(7)$ are false and why $P(3)$ and $P(5)$ are true.

Consider the following question:

For which natural numbers

n

do there exist nonnegative integers

x

and

y

such that

n = 3x + 5y

?

To help answer this question, we will let $\mathbb{Z}^* = \left\{ {x \in \mathbb{Z} \mid x \geq 0} \right\}$, and let $P \left( n \right)$ be There exist $x,y \in \mathbb{Z}^*$ such that $n = 3x + 5y$. Notice that $P(1)$ is false since if both $x$ and $y$ are zero, then $3x + 5y = 0$ and if either $x > 0$ or $y > 0$, then $3x + 5y \geq 3$. Also notice that $P(6)$ is true since $6 = 3 \cdot 2 + 5 \cdot 0$ and $P(8)$ is true since $8 = 3 \cdot 1 + 5 \cdot 1$.

Explain why $P(9)$, $P(10)$, $P(11)$, and $P(12)$ are true.

Complete the following.

For $n \in \mathbb{N}$ with $n \geq 2$, make a conjecture about a formula for the product $\left( {1 - \dfrac{1}{4}} \right) \left( {1 - \dfrac{1}{9}} \right) \left( {1 - \dfrac{1}{{16}}} \right) \cdots \left( {1 - \dfrac{1}{{n^2 }}} \right)$.

Let $y = \ln x$.

Determine $\dfrac{{dy}}{{dx}}$, $\dfrac{{d^2 y}}{{dx^2 }}$, $\dfrac{{d^3 y}}{{dx^3 }}$, and $\dfrac{{d^4 y}}{{dx^4 }}$.

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 $a_1 = 1$ 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 $a_1 = 1$, $a_2 = 3$, 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 $a_6$.