递归公式

递归可以由两个属性定义。基本情况和递归步骤。基本情况是不使用递归来产生结果的终止方案。递归步骤由一组规则组成，这些规则将连续案例减少到转发到基本案例。

递归公式

递归函数是使用前一项定义序列的每个项的函数，即，下一项取决于一个或多个已知的先前项。递归函数h(x) 写成——

h(x) = a ₀ h(0) + a ₁ h(1) + a ₂ h(2) + ... + a _{x – 1} h(x – 1)

其中 a _i ≥ 0 且 i = 0, 1, 2, 3, ... x – 1

递归公式是使用前项/前项定义序列的每个项的公式。它定义了以下参数

1. 序列的第一项
2. 从之前的术语中获取任何术语的模式规则。

很少有递归公式可以根据给定数据的模式找到第ⁿ项。他们是，

• n^th term of Arithmetic Progression a_n = a_{n – 1} + d for n ≥ 2
• n^th term of Geometric Progression a_n = a_{n – 1} × r for n ≥ 2
• n^th term in fibonacci Sequence a_n = a_{n – 1} + a_{n – 2} for n ≥ 2 and a₀ = 0 & a₁ = 1

Where d is common difference and r is the common ratio

示例问题

问题1：给定一系列数字，中间1、11、21、?、41有一个缺失的数字。使用递归公式找到缺失的项。

解决方案：

Given,

1, 11, 21, _, 41

First term (a) = 1

Difference between terms = 11 – 1 = 10

21 – 11 = 10

So the difference between numbers is same.

Common Difference (d) = 10

Recursive Function to find nth term is a_n = a_n-1 + d

a₄ = a_4-1 + d

= a₃ + d

= 21 + 10

a₄ = 31

Missing term in the given series is 31.

问题 2：给定数列 5, 9, 13, 17, 21,... 从给定数列中找出递归公式

解决方案：

Given number series

5, 9, 13, 17, 21,…

first term (a) = 5

Difference between terms = 9 – 5 = 4

13 – 9 = 4

17 – 13 = 4

21 – 17 = 4

So the difference between numbers is same.

Common Difference (d) = 4

The given number series is in Arithmetic progression.

So recursive formula a_n = a_n-1 + d

a_n = a_n-1 + 4

问题3：给定一系列数字，中间1、3、9、_、81、243有一个缺失的数字。使用递归公式找到缺失的项。

解决方案：

Given,

1, 3, 9, _, 81, 243

First term (a) = 1

The difference between numbers are large so It should not be in Arithmetic progression. 

Let’s check whether it is in Geometric progression or not,

a₂/a₁ = 3/1 = 3

a₃/a₂ = 9/3 = 3

a₅/a₄ = 243/81 = 3

Hence ratio between adjacent numbers are same. So given series is in Geometric Progression 

Common Ratio (r) = 3

Recursive Function to find nth term is a_n = a_n-1 × r

a₄ = a_4-1 × r

= a₃ × r

= 9 × 3

a₄ = 27

Missing term in the given series is 27.

问题 4：给定数列 2, 4, 8, 16, 32, ... 从给定数列中找出递归公式。

解决方案：

Given number series,

2, 4, 8, 16, 32, …

First term (a) = 2

Difference between terms = 4 – 2 = 2

8 – 4 = 4

It is not in A.P as difference between numbers are not same.

Let’s check whether it is in Geometric progression or not

a₂/a₁ = 4/2 = 2

a₃/a₂ = 8/4 = 2

a₄/a₃ = 16/8 = 2

Common Ratio (r) = 2

The given number series is in Geometric progression.

So recursive formula a_n = a_n-1 × r

a_n = a_n-1 × 2

问题 5：如果第³项和第 4^项分别为 2,3，则在^斐波那契数列中找到第 5 项。

解决方案：

Given that number series is in fibonacci series form.

Also given a₃ = 2

a₄ = 4

Then a₅ = a₃ + a₄

= 2 + 3

a₅ = 5

问题 6：在给定系列 1、1、2、3、5、8、13 中找到下一项

解决方案：

Given, 

1, 1, 2, 3, 5, 8, 13,…

The given series is in fibonacci form because every n^th term is the result of addition between two previous terms i.e., n – 1^th & n – 2^th terms

Example: a₃ = a₁ + a₂

3 = 1 + 2

Then a₈ = a₇ + a₆

= 13 + 8

a₈ = 21