前N个自然数的和系列的和

给定自然数n ，找到前N个自然数的和系列之和。

Sum-Series: is sum of first N natural numbers, i.e, sum-series of 5 is 15 ( 1 + 2 + 3 + 4 + 5 ).
 

例子：

Input: N = 5 
Output: 35 
Explanation: 
Sum of sum-series of {1, 2, 3, 4, 5} i.e. {1 + 3 + 6 + 10 + 15} is 35.
Input: N = 2 
Output: 4 
Explanation: 
Sum of sum-series of {1, 2} i.e. {1 + 3} is 4. 
 

简单方法：
查找从1到N的每个值的和系列，然后将其相加。

• 创建一个变量Total_sum来存储所需的求和序列。
• 从1到N遍历数字
• 通过使用公式sum =(N *(N + 1))/ 2来找到每个值的和系列
• 将值添加到Total_sum

最后，打印存储在Total_sum中的值。

C++

// C++ program to implement
// the above approach
#include
using namespace std;
 
// Function to find the sum
static long sumOfSumSeries(int N)
{
    long sum = 0L;
 
    // Calculate sum-series
    // for every natural number
    // and add them
    for (int i = 1; i <= N; i++)
    {
        sum = sum + (i * (i + 1)) / 2;
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int N = 5;
    cout << sumOfSumSeries(N);
}
 
// This code is contributed by Code_Mech

Java

// Java program to implement
// the above approach
 
class GFG {
 
    // Function to find the sum
    static long sumOfSumSeries(int N)
    {
 
        long sum = 0L;
 
        // Calculate sum-series
        // for every natural number
        // and add them
        for (int i = 1; i <= N; i++) {
            sum = sum + (i * (i + 1)) / 2;
        }
 
        return sum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 5;
        System.out.println(sumOfSumSeries(N));
    }
}

Python3

# Python3 program to implement
# the above approach
 
# Function to find the sum
def sumOfSumSeries(N):
 
    _sum = 0
 
    # Calculate sum-series
    # for every natural number
    # and add them
    for i in range(N + 1):
        _sum = _sum + (i * (i + 1)) // 2
 
    return _sum
 
# Driver code
N = 5
 
print(sumOfSumSeries(N))
     
# This code is contributed by divyamohan123

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to find the sum
static long sumOfSumSeries(int N)
{
    long sum = 0L;
 
    // Calculate sum-series
    // for every natural number
    // and add them
    for(int i = 1; i <= N; i++)
    {
       sum = sum + (i * (i + 1)) / 2;
    }
     
    return sum;
}
 
// Driver code
public static void Main()
{
    int N = 5;
     
    Console.Write(sumOfSumSeries(N));
}
}
 
// This code is contributed by Nidhi_Biet

Javascript

 
    // Javascript program to implement
    // the above approach
     
    // Function to find the sum
    function sumOfSumSeries(N)
    {
        let sum = 0;
 
        // Calculate sum-series
        // for every natural number
        // and add them
        for (let i = 1; i <= N; i++)
        {
            sum = sum + (i * (i + 1)) / 2;
        }
 
        return sum;
    }
     
    let N = 5;
    document.write(sumOfSumSeries(N));
     
    // This code is contributed by suresh07.

输出：

35

时间复杂度：O(N)
高效的方法：
可以使用以下公式直接计算以上系列的total_sum：

$\text{Total sum} = \frac{(N*(N+1)*(N+2))}{6}$
where N is the natural number 
 

以上公式的证明：
假设N = 5

1. 然后总和就是表中所有下面元素的总和，我们称其为“结果”

Natural number	1	2	3	4	5	6
Sum of natural number (sum-series)	1	3	6	10	15	21
Sum of sum-series	1	4	10	20	35	56

让我们在其他列中填充具有相同值的空单元格，将其称为“ totalSum ”

1
1	2
1	2	3
1	2	3	4
1	2	3	4	5

As sum of N numbers is repeated N times 
totalSum = N * [(N*(N + 1))/2]
populated data = (1 times * 2) + (2 times * 3) + (3 times * 4) + (4 times * 5) 
= 1*2 + 2*3 + 3*4 ……… +(N-1)*N 
=[(N-1) * (N) * (N+1)]/3 
 

1. 自从，

result = totalSum – populatedData 
= N * [(N*(N+1))/2] – [(N-1) * (N) * (N+1)]/3 
= (N*(N+1)*(N+2))/6 
 

2. 所以
   $\text{Sum of Sum-Series till N} = \frac{(N*(N+1)*(N+2))}{6}$

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
#include 
using namespace std;
 
// Function to find the sum
long sumOfSumSeries(int n)
{
    return (n * (n + 1) * (n + 2)) / 6;
}
 
// Driver code
int main ()
{
    int N = 5;
    cout << sumOfSumSeries(N);
    return 0;
}
 
// This code is contributed
// by shivanisinghss2110

Java

// Java program to implement
// the above approach
 
class GFG {
 
    // Function to find the sum
    static long sumOfSumSeries(int n)
    {
        return (n * (n + 1) * (n + 2)) / 6;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 5;
        System.out.println(sumOfSumSeries(N));
    }
}

Python3

# Python3 program to implement
# the above approach
 
# Function to find the sum
def sumOfSumSeries(n):
     
    return (n * (n + 1) * (n + 2)) // 6
 
# Driver code
N = 5
 
print(sumOfSumSeries(N))
 
# This code is contributed by divyamohan123

C＃

// C# program to implement the
// above approach
using System;
class GFG{
 
// Function to find the sum
static long sumOfSumSeries(int n)
{
    return (n * (n + 1) * (n + 2)) / 6;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 5;
     
    Console.Write(sumOfSumSeries(N));
}
}
 
// This code is contributed by Ritik Bansal

Java脚本

    // Javascript program to implement
    // the above approach
     
    // Function to find the sum
    function sumOfSumSeries(n)
    {
        return (n * (n + 1) * (n + 2)) / 6;
    }
     
    let N = 5;
    document.write(sumOfSumSeries(N));
 

输出：

35

考虑乘法，加法和除法的时间复杂度O(1)需要固定时间。