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

15 Jun, 2022
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++
Java
Python3
C#
PHP
Javascript
/* 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 seriesint seriesSum(int n){    int sum = 0;    for (int i=1; i<=n; i++)       sum += i*(i+1)/2;    return sum;} // Driver codeint main(){    int n = 4;    cout << seriesSum(n);    return 0;}
// 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 program to find sum# series 1, 3, 6, 10, 15, 21...# and then find its sum. # Function to find the sum of seriesdef seriessum(n):         sum = 0    for i in range(1, n + 1):        sum += i * (i + 1) / 2    return sum     # Driver coden = 4print(seriessum(n)) # This code is Contributed by Azkia Anam.
// 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 program to find sum// series 1, 3, 6, 10, 15, 21...// and then find its sum // Function to find// the sum of seriesfunction 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.?>
<script>// javascript program to find sum// series 1, 3, 6, 10, 15, 21...// and then find its sum // Function to find the sum of seriesfunction seriesSum(n){    let sum = 0;    for (let i = 1; i <= n; i++)       sum += i * ((i + 1) / 2);    return sum;} // Driver codelet 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 :  
It’s true for some n (n = 1, in this case).If it’s true for n, then it’s true for n+1.
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++
Java
Python3
C#
PHP
Javascript
/* 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 seriesint seriesSum(int n){    return (n * (n + 1) * (n + 2)) / 6;} // Driver codeint main(){    int n = 4;    cout << seriesSum(n);    return 0;}
// java program to find sum// series 1, 3, 6, 10, 15, 21...// and then find its sumimport 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
# Python 3 program to find sum# series 1, 3, 6, 10, 15, 21...# and then find its sum*/ # Function to find the sum of seriesdef seriesSum(n):     return int((n * (n + 1) * (n + 2)) / 6)  # Driver coden = 4print(seriesSum(n)) # This code is contributed by Smitha.
// C# program to find sum// series 1, 3, 6, 10, 15, 21...// and then find its sumusing 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 program to find sum// series 1, 3, 6, 10, 15, 21...// and then find its sum // Function to find// the sum of seriesfunction seriesSum($n){    return ($n * ($n + 1) *           ($n + 2)) / 6;} // Driver code$n = 4;echo(seriesSum($n)); // This code is contributed by Ajit.?>
<script>/* javascript program to find sum series 1, 3, 6, 10, 15, 21...and then find its sum*/  // Function to find the sum of seriesfunction 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)
Smitha Dinesh Semwal
jit_t
aashish1995
todaysgaurav
polymatir3j
series
triangle
triangular-number
Mathematical
Mathematical
series
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