几何系列

一组按顺序排列的事物称为序列，当序列开始遵循某种模式时，它们称为Progressions 。渐进具有不同类型，例如算术渐进，几何渐进，和声渐进。

特定Sequence的总和称为Series 。一个序列可以是无限的或有限的，具体取决于序列，如果序列是无限的，它将给出无限的级数，而如果序列是有限的，它将给出有限的级数。

Let’s take a finite Sequence:

a_1, a_2, a_3, a_4, a_5,……….a_n

The Series of this Sequence is given as:

a₁+ a₂+ a₃+ a₄+a₅+……….a_n

The Series is also denoted as :

$\sum_{k=1}^{n}a_k$

Series is represented using Sigma (∑) Notation in order to Indicate Summation.

几何系列

在几何级数中，每个下一项都是前项乘以某个常数的乘积，并且取决于该常数的值，该级数可能会增加或减少。

Geometric Sequence is given as: 

a, ar, ar², ar³, ar⁴,….. {Infinite Sequence}

a, ar, ar², ar³, ar⁴, ……. arⁿ {Finite Sequence}

Geometric Series for the above is written as:

a + ar + ar² + ar³ + ar⁴ +…. {Infinite Series}

a + ar + ar² + ar³ + ar⁴ +….. arⁿ {Finite Series}

Where. a = First term

r = Common Factor

‘a’和’r’的值可以为0吗?

Answer: No, the value of a≠0, if the first term becomes zero, the series will not continue. Similarly, r≠0.

几何级数公式

The Geometric Series formula for the Finite series is given as,

${S_n =\frac{a(1-r^n)}{1-r}}$

where, S_n = sum up to n^th term

a = First term

r = common factor

几何级数公式的推导

Suppose a Geometric Series for n terms: 

S_n = a + ar + ar² + ar³ + …. + ar^n-1 ⇢ (1)

Multiplying both sides by the common factor (r):

r S_n = ar + ar² + ar³ + ar⁴ + … + arⁿ ⇢ (2)

Subtracting Equation (1) from Equation (2):

(r S_n – S_n) = (ar + ar² + ar³ + ar⁴ + … arⁿ) – (a + ar + ar² + ar³ +… + ar^n-1)

S_n (r-1) = arⁿ – a

S_n (1 – r) = a (1-rⁿ)

${S_n =\frac{a(1-r^n)}{1-r}}$

Note: When the value of k starts from ‘m’, the formula will change.

$\sum_{k=m}^{n}ar^k=\frac{a(r^m-r^{n+1}}{1-r}$, when r≠0

对于无限几何级数

n will tend to Infinity, n⇢∞, Putting this in the generalized formula:

$S_\infty = \sum_{n=1}^{\infty}ar^{n-1} = \frac{a}{1-r}; -1<{r}<1$

N^th term for the G.P. : a_n = ar^n-1

几何系列产品

几何级数中所有数字的乘积为我们提供整体乘积。在计算整个序列的几何平均值时非常有用。

几何平均数

根据定义，它是n^个数字乘积的第n个根，其中“ n”表示序列中存在的项数。几何均值不同于算术均值，后者是通过将所有项相加并除以“ n”而获得的，而前者是通过乘积然后取所有项的均值来获得的。

几何均值的意义

计算几何平均值是因为它可以告知不同时期之间发生的复利。它通过采用几何级数的平均值来告诉级数的中心行为。例如，可以使用几何平均值轻松地分析细菌的生长。简而言之，“时间范围越长”或系列中的值彼此不同，复合变得越关键，因此“几何平均值”更适合使用。

几何均值公式

$\bar{X}_{geom}=\sqrt [n]{x_{1}.x_{2}.x_{3}...x_{n}}$

where,

$\bar{X}_{geom}=\text{symbol for Geometric mean}\\x_{1},x_{2},x_{3}...x_{n}=\text{terms present in the geometric series}\\n=\text{number of terms present in the series}$

问题1：什么是几何平均值2、4、8?

回答：

According to the formula, 

$=\sqrt [3]{(2)(4)(8)}\\=4$

问题2：在以下几何级数中找到第一个项和公因数：

4、8、16、32、64等。

回答：

Here, It is clear that the first term is 4, a=4

We obtain common Ratio by dividing 1st term from 2nd:

r = 8/4 = 2

问题3：查找^第8和^第n项为GP：3，9，27，81，…。

答：

Put n=8 for 8^th term in the formula: ar^n-1

For the G.P : 3, 9, 27, 81….

First term (a) = 3

Common Ratio (r) = 9/3 = 3

8^th term = 3(3)^8-1 = 3(3)⁷ = 6561

N^th = 3(3)^n-1 = 3(3)ⁿ(3)^-1

= 3ⁿ

问题4：对于GP：2、8、32…。哪个术语的值为131073?

回答：

Assume that the value 131073 is the N^th term,

a = 2, r = 8/2 = 4

N^th term (a_n) = 2(4)^n-1 = 131073

4^n-1 = 131073/2 = 65536

4^n-1 = 65536 = 4⁸

Equating the Powers since the base is same:

n-1 = 8

n = 9

问题5：求出该序列的^第5个和第N个项的总和： $1, \frac{1}{2},\frac{1}{4},\frac{1}{8}...$

回答：

a= 1, r = 1/2

Sum of N terms for the G.P, ${S_n =\frac{a(1-r^n)}{1-r}}$                                     

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

 $= 2 (1-(\frac{1}{2})^n)$

Sum of first 5 terms ⇒ a₅ = $2 ( 1-(\frac{1}{2})^5)$

= $2 ( 1-(\frac{1}{32}))$

= $(\frac{31}{16})$

问题6：求出无限GP的总和：0.5、1、2、4、8，…

回答：

Formula for the Sum of Infinite G.P: $\frac{a}{1-r}$ ; r≠0

a = 0.5, r = 2

S_∞= (0.5)/(1-2) = 0.5/(-1)= -0.5

问题7：求系列的总和：5、55、555、5555，…n个术语

答：

The given Series is not in G.P but it can easily be converted into a G.P with some simple modifications.

Taking 5 common from the series: 5 (1, 11, 111, 1111,… n terms)

Dividing and Multiplying with 9: $\frac{5}{9}(9+ 99+ 999+...n terms)$

⇒ $\frac{5}{9}[((10+(10)^2+(10)^3+...n terms)-(1+1+1+...n terms)]$

⇒$\frac{5}{9}[(\frac{10((10)^n-1)}{10-1})-(n)]$

⇒ $\frac{5}{9}[(\frac{10((10)^n-1)}{9})-(n)]$