Sum of the series 1, 3, 6, 10... (Triangular Numbers)

Sum of the series 1, 3, 6, 10… (Triangular Numbers)

Last Updated : 31 Oct, 2018

Given n, no of elements in the series, find the summation of the series 1, 3, 6, 10….n. The series mainly represents triangular numbers.

Examples:

Input: 2
Output: 4
Explanation: 1 + 3 = 4

Input: 4
Output: 20
Explanation: 1 + 3 + 6 + 10 = 20

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to one by one add triangular numbers.

C++

/* CPP program to find sum 
 series 1, 3, 6, 10, 15, 21... 
and then find its sum*/
#include <iostream> 
using namespace std; 
  
// Function to find the sum of series 
int seriesSum(int n) 
{ 
    int sum = 0; 
    for (int i=1; i<=n; i++) 
       sum += i*(i+1)/2; 
    return sum; 
} 
  
// Driver code 
int main() 
{ 
    int n = 4; 
    cout << seriesSum(n); 
    return 0; 
} 

Java

// Java program to find sum 
// series 1, 3, 6, 10, 15, 21... 
// and then find its sum*/ 
import java.io.*; 
  
class GFG { 
          
    // Function to find the sum of series 
    static int seriesSum(int n) 
    { 
        int sum = 0; 
        for (int i = 1; i <= n; i++) 
        sum += i * (i + 1) / 2; 
        return sum; 
    } 
  
    // Driver code 
    public static void main (String[] args)  
    { 
        int n = 4; 
        System.out.println(seriesSum(n)); 
          
    } 
} 
  
// This article is contributed by vt_m 

Python3

# Python3 program to find sum 
# series 1, 3, 6, 10, 15, 21... 
# and then find its sum. 
  
# Function to find the sum of series 
def seriessum(n): 
      
    sum = 0
    for i in range(1, n + 1): 
        sum += i * (i + 1) / 2
    return sum
      
# Driver code 
n = 4
print(seriessum(n)) 
  
# This code is Contributed by Azkia Anam. 

C#

// C# program to find sum 
// series 1, 3, 6, 10, 15, 21... 
// and then find its sum*/ 
using System; 
  
class GFG { 
  
    // Function to find the sum of series 
    static int seriesSum(int n) 
    { 
        int sum = 0; 
          
        for (int i = 1; i <= n; i++) 
            sum += i * (i + 1) / 2; 
              
        return sum; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int n = 4; 
          
        Console.WriteLine(seriesSum(n)); 
    } 
} 
  
// This article is contributed by vt_m. 

PHP

<?php 
// PHP program to find sum 
// series 1, 3, 6, 10, 15, 21... 
// and then find its sum 
  
// Function to find  
// the sum of series 
function seriesSum($n) 
{ 
    $sum = 0; 
    for ($i = 1; $i <= $n; $i++) 
        $sum += $i * ($i + 1) / 2; 
    return $sum; 
} 
  
// Driver code 
$n = 4; 
echo(seriesSum($n)); 
  
// This code is contributed by Ajit. 
?> 

Output:

Time complexity : O(n)

An efficient solution is to use direct formula n(n+1)(n+2)/6

Let g(i) be i-th triangular number.
g(1) = 1
g(2) = 3
g(3) = 6
g(n) = n(n+1)/2

Let f(n) be the sum of the triangular
numbers 1 through n.
f(n) = g(1) + g(2) + ... + g(n)

Then:
f(n) = n(n+1)(n+2)/6

How can we prove this? We can prove it by induction. That is, prove two things :

It’s true for some n (n = 1, in this case).
If it’s true for n, then it’s true for n+1.

This allows us to conclude that it’s true for all n >= 1.

Now 1) is easy. We know that f(1) = g(1) 
= 1. So it's true for n = 1.

Now for 2). Suppose it's true for n. 
Consider f(n+1). We have:
f(n+1) = g(1) + g(2) + ... + g(n) + g(n+1) 
       = f(n) + g(n+1)

Using our assumption f(n) = n(n+1)(n+2)/6 
and g(n+1) = (n+1)(n+2)/2, we have:
f(n+1) = n(n+1)(n+2)/6 + (n+1)(n+2)/2
       = n(n+1)(n+2)/6 + 3(n+1)(n+2)/6
       = (n+1)(n+2)(n+3)/6
Therefore, f(n) = n(n+1)(n+2)/6

Below is the implementation of the above approach:

C++

/* CPP program to find sum 
 series 1, 3, 6, 10, 15, 21... 
and then find its sum*/
#include <iostream> 
using namespace std; 
  
// Function to find the sum of series 
int seriesSum(int n) 
{ 
    return (n * (n + 1) * (n + 2)) / 6;  
} 
  
// Driver code 
int main() 
{ 
    int n = 4; 
    cout << seriesSum(n); 
    return 0; 
} 

Java

// java program to find sum 
// series 1, 3, 6, 10, 15, 21... 
// and then find its sum 
import java.io.*; 
  
class GFG  
{ 
    // Function to find the sum of series 
    static int seriesSum(int n) 
    { 
        return (n * (n + 1) * (n + 2)) / 6;  
    } 
  
   // Driver code 
    public static void main (String[] args) { 
          
        int n = 4; 
        System.out.println( seriesSum(n)); 
          
    } 
} 
  
// This article is contributed by vt_m 

Python3

# Python 3 program to find sum 
# series 1, 3, 6, 10, 15, 21... 
# and then find its sum*/ 
  
# Function to find the sum of series 
def seriesSum(n): 
  
    return int((n * (n + 1) * (n + 2)) / 6) 
  
  
# Driver code 
n = 4
print(seriesSum(n)) 
  
# This code is contributed by Smitha. 

C#

// C# program to find sum 
// series 1, 3, 6, 10, 15, 21... 
// and then find its sum 
using System; 
  
class GFG { 
      
    // Function to find the sum of series 
    static int seriesSum(int n) 
    { 
        return (n * (n + 1) * (n + 2)) / 6; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
  
        int n = 4; 
          
        Console.WriteLine(seriesSum(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find sum 
// series 1, 3, 6, 10, 15, 21... 
// and then find its sum 
  
// Function to find  
// the sum of series 
function seriesSum($n) 
{ 
    return ($n * ($n + 1) *  
           ($n + 2)) / 6;  
} 
  
// Driver code 
$n = 4; 
echo(seriesSum($n)); 
  
// This code is contributed by Ajit. 
?> 

Output:

Time complexity : O(1)

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Related Articles

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP