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# 数学代写代考| Arithmetic and Geometric Series 离散数学

## 数学代写| Arithmetic and Geometric Series 离散代考

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## 离散数学代写

An arithmetic series is the sum of the terms in an arithmetic sequence, and geometric sequence is the sum of the terms in a geometric sequence. It is possible t derive a simple formula for the sum of the first $n$ terms in arithmetic and geometri series.
Arithmetic Series
We write the series in two ways: first, the normal left to right addition, and then th reverse, and then we add both series together:
\begin{aligned} &S n=a+(a+d)+(a+2 d)+(a+3 d)+\cdots+(a+(n-1) d) \ &S n=a+(n-1) d+a+(n-2) d+\cdots++(a+d)+a \end{aligned}
$S n=a+(a+d)+(a+2 d)+(a+3 d)+\cdots+(a+(n-1) d)$ $S n=a+(n-1) d+a+(n-2) d+\cdots+\quad+(a+d)+a$ $\cdots \cdot \cdots+(2 a+(n-1) d]+\cdots+[2 a+(n-1) d] \quad(n$ times $)$ $2 S n=[2 a+(n-1) d]+[(n-1) d]$ $2 S n=n \times[2 a+(-\cdots$ $2 S n=n \times[2 a+(n-1) d] .$
Therefore, we conclude that (where $r \neq 1$ )
$$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}=\frac{a\left(1-r^{n}\right)}{1-r} .$$
The case when $r=1$ corresponds to the arithmetic series $a+a+\cdots+a$, an the sum of this series is simply na. The geometric series converges when $|r|<1$ a $r^{n} \rightarrow 0$ as $n \rightarrow \propto$, and so
$$S_{n} \rightarrow \frac{a}{1-r} \quad \text { as } n \rightarrow \infty$$
Example (Geometric Series)
Find the sum of the first $n$ terms in the following geometric series $1,1 / 2,1 / 4,1 / 8, \ldots$ What is the sum of the series?
Solution
Clearly, $a=1$ and $r=1 / 2$. Therefore, applying the formula, we get

$$\开始{对齐} &S n=a+(a+d)+(a+2 d)+(a+3 d)+\cdots+(a+(n-1) d) \ &S n=a+(n-1) d+a+(n-2) d+\cdots++(a+d)+a \end{对齐}$$
$S n=a+(a+d)+(a+2 d)+(a+3 d)+\cdots+(a+(n-1) d)$ $S n=a+(n-1) d+a+ (n-2) d+\cdots+\quad+(a+d)+a$ $\cdots \cdot \cdots+(2 a+(n-1) d]+\cdots+[2 a+(n-1) d] \quad (n$ 次 $)$ $2 S n=[2 a+(n-1) d]+[(n-1) d]$ $2 S n=n \times[2 a+(-\cdots$ $2 S n= n \times[2 a+(n-1) d] .$

$$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}=\frac{a\left(1-r^{n}\right)}{1- r} 。$$
$r=1$ 对应算术级数$a+a+\cdots+a$ 的情况，这个级数的和就是na。当 $|r|<1$ a $r^{n} \rightarrow 0$ as $n \rightarrow \propto$ 时，几何级数收敛，依此类推
$$S_{n} \rightarrow \frac{a}{1-r} \quad \text { as } n \rightarrow \infty$$

## 图论代考

$n=0$ 或 $n=1$，而递归步骤指定如何应用

$n(n-1)(n-2) \ldots 3.2 .1$ 和 $0 !=1$。根据基本情况的正式定义

$\begin{array}{ll}\text { 基本步骤 } & \text { fac }(0)=1 \ \text { 递归步骤 } & \text { fac }(n)=n * \operatorname{fac }(n-1)\end{数组}$

$n=0$ 或 $n=1$，而递归步骤指定如何应用

$\mathrm{~ T h e ~ r e a d e r ~ i s ~ r e f e r}$

## 密码学代考

• Cryptosystem
• A system that describes how to encrypt or decrypt messages
• Plaintext
• Message in its original form
• Ciphertext
• Message in its encrypted form
• Cryptographer
• Invents encryption algorithms
• Cryptanalyst
• Breaks encryption algorithms or implementations

## 编码理论代写

1. 数据压缩（或信源编码
2. 前向错误更正（或信道编码
3. 加密编码
4. 线路码