评估系列： (a) sum_{k=1}^{20) k (b) sum_{j = 1}^{60} 5 j^2 (c) sum_{i = 1}^{11} 2ln3- 5i

数字系统是使用数字（0-9）和一组原始规则在定义的数字线上表示数字的概念。数字系统可以有多种类型，例如，

1. 十进制数系统（Base – 10 ，使用数字 (0 – 9)）
2. 二进制数字系统（Base – 2，使用数字 0 和 1）
3. 八进制数系统（Base – 8，使用数字 (0 – 7)
4. 十六进制数系统（Base-16，使用数字 (0 – 9) 和字母 (A – F)

序列和系列

数字系列是数字的序列/系列。在关于数列的问题中，可以有一种模式或规则应用于数字以获得给定的数列。因此，数列可以是数列的乘积，也可以是数列的总和。例子：

1. 前 10 个自然数之和
2. 前 10 个偶数之和
3. 前 10 个奇数之和
4. 前 10 个自然数的乘积。

这些序列可以通过使用各种技术来解决，例如算术级数 (AP)、几何级数 (GP)、算术-几何级数 (AGP)、最小二乘估计、回归算法等，但解决序列的最基本技术是算术级数（AP）和算术-几何级数（AGP）。下面可以看到一些算术级数和几何级数公式。

算术级数公式

a_r = a₁ + (r – 1)d

S_n = n/2(2a + (n – 1)d)

n – number of elements in a series

d – common difference between two consecutive elements

a₁ – first element in the series

a_r – rth element in the series

Common difference (d) = a₂ – a₁

几何级数

a_i  = a₁r^{i – 1}

S_n =a (rⁿ – 1)/(r – 1)

n – number of elements in the series

r – common ratio between two consecutive elements

a₁ – first element in the series

a_i – ith element in the series

r = a₂/a₁

评估系列：（a）给定的总和， $\sum_{k=1}^{20} k$ (二) $\sum_{j = 1}^{60} 5 j^2$ （C） $\sum_{i = 1}^{11} 2 ln3-5i$

解决方案：

(a) The given sum,$\sum_{k=1}^{20} k$ can be illustrated as,-

1 + 2 + 3 + …. + 20

The above series is a sum of the first 20 natural numbers and the consecutive numbers have a common difference (d) = 1 and the first number (a₁) = 1. Now that the series is in a form of Arithmetic Progression (AP), where, sum of n terms in AP can be given as, 

$(S_n) = \frac{n}{2}(2a_1+(n-1)d)$

By substituting, n = 20, a₁= 1, d=1. 

 $S_{20}= \frac{20}{2}(2\times1+(20-1)\times1)=10 \times21 =210$ 

(b) $\sum_{j = 1}^{60} 5 j^2$

The above sum is 5 times the sum of squares of the first 60 natural numbers and it can be illustrated as, 

5( 1² + 2² + …. + 60²)

The sum of squares for n- terms looks like$S_n=1^2+2^2+\dots+n^2$ , whose sum can be solved using $S_n= \frac{n(n+1)(2n+1)}{6}$

So, given sum(S) can be written as,

S = 5 S₆₀

$S = 5 \times\frac{(60)(61)(121)}{6} \\ S= 5 \times 10\times61\times121 = 369050$

(c) $\sum_{i = 1}^{11} 2 ln3-5 i$

The series can be seen as a combination of a constant component and a variable. Such series must be decomposed into constant and variable components to solve the series easily and the above sum can be illustrated as,

$2ln3-5(1) +2ln3-5(2)+\dots+2ln3-5(11)$

Adding up all the elements,

$S = 2ln3\underbrace{(1+1+\dots+1)}_\textit{11 times} - 5(1+2+3+ \dots+11)$

Sum of n-terms in AP with common difference d =1 can be written as $S = \frac{n(n+1)}{2}$ , where n = 11, as in the query.

$S= 2ln3(11) - 5 \frac{(11)(12)}{2}$

S = 22ln3 – 330 

S = -305.83

类似问题

问题1：解决系列， $\sum_{r=1}^{10} r^2 + r$

解决方案 ：

The above series can be illustrated as follows,

(1²+2²+ …+10² ) + (1 + 2+… 10) 

This series can be decomposed into two independent series, where one series is the sum of squares of the first 10 natural numbers and the other series would be the sum of the first natural numbers. 

S = S₁ + S₂

S₁ = n(n + 1)(2n + 1)/6 , sum of squares of first n natural numbers

S₂ = n(n + 1)/2 ,  sum of first n natural numbers

By substituting n = 10 , S₁ = 385 and S₂ = 55, so 
S = 385 + 55

S = 440 

问题2：解决系列， $\sum_{r=1}^{20}r^3 + r^2$

解决方案 ：

This series can be decomposed into two independent series just like the above question, where one series is the sum of cubes of the first 20 natural numbers and the other series is the sum of squares of the first 20 natural numbers. Sum (S) can be written as the sum of the independent series.

S = (1³ + 2³ + 3³ + … + 20³ ) + (1² + 2² + … + 20²)

S = S₁ + S₂

S₁ = n²(n + 1)²/4 , sum of cubes of first n natural numbers

S₂ = n(n + 1)(2n + 1)/6 , sum of squares of first n natural numbers

By substituting n = 20, S₁ = 44100 and S₂ = 2870

As said, S is the composition of S₁ and S₂, So,

S = S₁ + S₂ = 46970

问题3：解决系列， $\sum_{r=0}^{10} 2^r$

解决方案 ：

The series can be illustrated as,

S = 1 + 2 + 2² + 2³ + … 2¹⁰ 

The above series follows geometric progression with common ratio (r) = 2/1 = 2 , a₁ = 1, a_n = 2¹⁰, n = 11 elements

S = a₁ (rⁿ – 1) / (r – 1) , S -> Sum of elements in GP 

By substituting a₁ = 1 , r = 2, n = 11

S = 1 (2¹¹ -1) /(2 – 1) 

S = 2¹¹ – 1 .