几何序列和级数

序列定义为按特定顺序排列的数字，即数字的有序列表。例如：1、3、5、7，…等。

序列有2种类型：

Arithmetic sequence: An arithmetic sequence is the one in which the difference between two consecutive terms is constant. This difference is known as common difference. 

Geometric sequence: In contrast, the geometric sequence is the one in which the ratio between two consecutive terms is constant. This ratio is known as common ratio.

系列

序列定义为序列元素的总和。例如：1 + 4 + 7 + 10 +…等。

系列有两种类型：

Finite Series: A finite series is one in which the number of elements in the series is known.

Infinite Series: When the number of elements in the series is not known, i.e. series with an infinite number of elements is known as infinite series. 

几何序列

几何序列是两个连续项之间的比率恒定的序列。该比率称为“ r”表示的公共比率，其中r≠0。

让序列的元素表示为：

 a₁, a₂, a_3, a₄, …, a_n

如果满足以下条件，则给定序列是几何序列：

a₁/a₂ = a₂/a₃ = a₃/a₄ = … = a_n-1/a_n = r (common ratio)

给定的序列也可以写成：

a, ar, ar², ar³, … , ar^n-1  

Here, r is the common ratio and a is the scale factor

共同比率由下式给出：

r = successive term/preceding term = ar^n-1 / ar^n-2

^{几何序列的第}N个术语是什么?

为了找到几何序列的第n个项，我们知道该序列的形式为a，ar，ar ² ，ar ³ ，ar ⁴ …………。

所述^第术语n由_正表示_。因此，找到几何序列的第n个项将是：

a_n = ar^n-1

公式的推导

给定GP的每一项为₁ ，a ₂ ，a ₃ ，a ₄ ，…，a _n ，根据第一项a ₁表示所有这些项，我们得到

a₁ = a₁  

a₂ = a₁r

a₃ = a₂r = (a₁r)r = a₁r²

a₄ = a₃r = (a₁r²)r = a₁r³

…

a_m = a₁r^m−1

…

a_n = a₁r^{n – 1}

在哪里，

a₁ = the first term, a₂ = the second term, and so on

a_n: the last term (or the nth term) and

a_m: any term before the last term

上一项的第n个项由下式给出：

a_n = l/r^n-1

where l is the last term

几何序列的前n个项的总和是多少?

几何序列的前n个项之和由下式给出：

S_n = a(1 – rⁿ)/(1 – r), if r < 1

S_n = a(rⁿ -1)/(r – 1), if r > 1

公式的推导

几何级数的总和(称为几何级数)由下式给出：

S = a₁ + a₂ + a₃ + … + a_n

S = a₁ + a₁r + a₁r² + a₁r³ + … + a₁rⁿ⁻¹     ….Equation (1)

Multiply both sides of Equation (1) by r (common ratio), we get

S × r= a₁r + a₁r² + a₁r³ + a₁r⁴ + … + a₁rⁿ     ….Equation (2)

Subtract Equation (2) from Equation (1)

S – Sr = a₁ – a₁rⁿ

(1 – r)S = a₁(1 – rⁿ)

S_n = a₁(1 – rⁿ)/(1 – r), if r<1

Now, Subtracting Equation (1) from Equation (2) will give

Sr – S = a₁rⁿ – a₁

(r – 1)S = a₁(rⁿ-1)

因此，

如果r> 1，则S _n = a ₁ (r ⁿ -1)/(r – 1)

无限项之和

无限几何级数中的项数将接近无穷大(n =∞)。无限的几何级数之和只能在| r |的范围内定义。 <1。

S = a(1 – rⁿ)/(1 – r)

S = (a – arⁿ)/(1 – r)

S = a/(1 – r) – arⁿ/(1 – r)

For n -> ∞, the quantity (arⁿ) / (1 – r) → 0 for |r| < 1, 

因此，

_S∞ = a /(1-r)，其中| r | <1

问题1：找到序列的公共比例和比例因子：4、12、36、108、324等

解决方案：

Sequence provided is 4, 12, 36, 108, 324, …

Common ratio = 12/4 = 3 

Scale factor = 4 

问题2：找到序列的公共比例和比例因子：5，-5、5，-5、5、5，…

解决方案：

Given Sequence, 5, -5, 5, -5, 5, -5, … 

Common ratio = -5/5 = -1 

Scale factor = 5

问题3：找到序列的n个项和n个项的总和：1、2、4、8、16、32

解决方案：

Given Sequence, 1, 2, 4, 8, 16, 32

Common ratio r = 2/1 = 2

Scale factor = 1

6th term in the sequence = ar^n-1 = 1.2^6-1 = 32

3rd term form last = l/r^n-1 = l/2^3-1 = 32/4 = 8

Sum of first 3 terms = a(rⁿ -1)/(r – 1) = 1(2³-1)/(2-1) = 7

几何级数的性质

• a ² _k = a _k-1 * a _{k + 1}
• a ₁ * a _n = a ₂ * a _n-1 =…= a _k * a _{n-k + 1}
• 如果我们将非零数量乘以或除以GP的每个项，则结果
  GP中的序列也具有相同的共同点。
• GP中所有术语的倒数也构成一个GP。
• 如果GP中的所有术语都提高到相同的幂，则新系列也将出现在GP中。
• 如果y ² = xz，则三个非零项x，y和z在GP中。

明确的公式

显式公式是一种定义相对于术语编号的序列术语的公式。几何序列的第n个项由显式公式给出：

a _n = a ₁ * r ^n-1

问题：给定₁ = 3和₄ = 24_{的几何序列，找到5}

解决方案：

The sequence can be written in terms of the initial term and the common ratio r.

Write the fourth term of sequence in terms of  a₁ and r.  Substitute 24 for a_4. Solve for the common ratio.

a_n = a₁ * r^n-1

a₄ = 3r³

24 = 3r³

8 = r³

r = 2

Find the second term by multiplying the first term by the common ratio.

a₅ = a₁ * r^n-1

= 3 * 2^5-1

= 3 * 16 = 48

递归公式

递归公式定义相对于先前值的序列项。与显式公式相反，后者相对于术语数字对其进行定义。

作为一个简单的例子，让我们看一下顺序：1、2、4、8、16、32

模式是重复乘以2。所以递归公式是

项(n)=项(n – 1)* 2

请注意，要查找任何术语，您必须知道前一个。每个项是公比与上一个项的乘积。

项(n)=项(n – 1)* r

问题：为以下几何序列编写一个递归公式：8，12，18，27，…

解决方案：

The first term is given as 6. The common ratio can be found by dividing the second term by the first term.

r = 12/8 = 1.5

Substitute the common ratio into the recursive formula for geometric sequences and define  a₁

term(n) = term(n – 1) * r 

= term(n -1) * 1.5 for n>=2

a₁ = 6

转换的几何序列形式

Explicit form: a_n = k * r^n-1

Recursive form: a₁ = k , a_n = a_n-1 * r

问题1：给定f(n)的递归公式：

f(1)= 6

f(n)= f(n-1)*(-6.5)

找出f(n)的明确公式

解决方案：

From the recursive formula, we can tell that the first term of the sequence is 6 and the common ratio is -6.5

Explicit formula : f(n) = 6 * (-6.5)^n-1

问题2：给定f(n)的显式公式：

f(n)= 6 *(-6.5) ^n-1

求f(n)的递归公式。

解决方案：

From the explicit formula, we can tell that the first term of the sequence is 6 and the common ratio is -6.5

Recursive formula: f(1) = 6

f(n) = f(n-1) * (-6.5)