前 1000 个自然数之和比前 1001 个自然数之和小多少？

数字是用于测量或计算数量的数学值或数字。它也用数字表示，例如 3、4、8 等。其他类型的数字包括整数、整数、自然数、有理数和无理数等。数字系统是一种包含多种数字的系统，例如素数、奇数、偶数、有理数、整数等。这些数字可以用数字和文字的形式来表达。例如，数字 55 和 75，当表示为数字时，也可以写成 55 和 75。

自然数

自然数是数字系统的一个子集，包括从 1 到无穷大的所有正整数。自然数有时被称为计数数，因为它们不包括零或负数。它们是实数的子集，仅包括正整数，不包括零、分数、小数和负数。只有正整数，即 1, 2, 3, 4, 5, 6,…，才包含在自然数集中。

自然数之和

n个自然数之和是一个算术级数，其中n项之和以第一项为1，n为项数，第n项为第n项的顺序排列。自然数是那些以一开始并以无穷大结束的数。除了数字 0，自然数只包含整数。 n个自然数之和，

[n(n+1)]/2 

示例：求前 30 个自然数之和。

解决方案：

Given, n = 30

The sum of natural numbers formula is:

S = [30(30 + 1)]/2

S = [30(30 + 1)]/2

S = [30(31)]/2

= 930 /2

= 465

前 1000 个自然数之和比前 1001 个自然数之和小多少？

解决方案：

To find the difference between both the sum of natural numbers,

For first 1000 natural numbers, [n(n + 1)]/2

 Here n = 1000 

= [1000(1000 + 1)]/2

= [1000(1001)]/2

= 1001000/2

= 500500

Sum of first 1000 natural numbers (S₁) are 500500

Now sum of first 1001 natural numbers are [n(n + 1)]/2 here n = 1001

= [1001(1001 + 1)]/2

= [1001(1002)]/2

= 1003002 / 2

= 501501

Sum of first 1001 natural numbers (S₂) are 501501

Now S₂ – S₁ = 501501 – 500500

= 1001

The sum of the first 1000 natural numbers is smaller by 1001 than the sum of the first 1001 natural numbers.

类似问题

问题1：求前100个自然数之和？

解决方案：

Here n = 100 

Therefore sum of 100 natural numbers are: [n(n + 1)]/2

= [100(100 + 1)]/2

= [100(101)]/2

= 10100/2

= 5050

问题 2：求前 100 个偶数自然数之和？

解决方案：

First 100 natural numbers,

As we know there are 100 even natural numbers from 1 to 200.

Therefore n = 100 

By using the formula of sum of even numbers S_n = n (n + 1)

S_n = 100 (100 + 1)

= 100 × 101

= 10100

问题 3：求前 200 个偶数自然数之和？

解决方案：

First 200 natural numbers,

As we know there are 200 even natural numbers from 1 to 400.

Therefore n=200

By using the formula of sum of even numbers S_n = n (n + 1)

S_n = 200 (200 + 1)

= 200 × 201

= 40200

问题 4：求前 50 个奇数自然数之和？

解决方案：

First 50 Odd natural numbers,

As we know there are 50 odd natural numbers from 1 to 100.

Therefore n = 50

By using the formula of sum of even numbers S_n = n²

S_n = n²

= 50²

= 2500

问题 5：求前 25 个奇数自然数之和？

解决方案：

First 25 Odd natural numbers,

As we know there are 25 odd natural numbers from 1 to 50.

Therefore n = 25

By using the formula of sum of even numbers S_n = n²

S_n = n²

= 25²

= 625

问题6：前100个自然数之和比前101个自然数之和小多少？

解决方案：

To find the difference between both the sum of natural numbers,

For first 100 natural numbers [n(n + 1)]/2

Here n = 100

= [100(100 + 1)]/2

= [100(101)]/2

= 10100/2

= 5050

Sum of first 1000 natural numbers (S₁) are 5050

Now sum of first 101 natural numbers are [n(n + 1)]/2, here n = 101

= [101(101 + 1)]/2

= [101(102)]/2

= 10302 / 2

= 5151

Sum of first 101 natural numbers (S₂) are 5151

Now S₂ – S₁  = 5151 – 5050

= 101

The sum of the first 100 natural numbers is smaller by 101 than the sum of the first 101 natural numbers.