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

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

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.
?>

Javascript

<script>
// javascript 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)
{
    let sum = 0;
    for (let i = 1; i <= n; i++)
       sum += i * ((i + 1) / 2);
    return sum;
}

// Driver code
let n = 4;
  document.write(seriesSum(n)) ;

// This code is contributed by aashish1995

</script>

Output: 

20

Time complexity : O(n)

Auxiliary Space: O(1) since using constant variables

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 : 

1. It’s true for some n (n = 1, in this case).
2. 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.
?>

Javascript

<script>
/* javascript 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
    let n = 4;
    document.write(seriesSum(n));

// This code is contributed by todaysgaurav

</script>

Output: 

20

Time complexity : O(1)
Auxiliary Space: O(1), since no extra space has been taken.