Given a positive integer n. The task is to find the sum of the sum of first n natural number.
Examples:
Input: n = 3
Output: 10
Explanation:
Sum of first natural number: 1
Sum of first and second natural number: 1 + 2 = 3
Sum of first, second and third natural number = 1 + 2 + 3 = 6
Sum of sum of first three natural number = 1 + 3 + 6 = 10Input: n = 2
Output: 4
A simple solution is to one by one add triangular numbers.
/* 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 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 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# 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));
}
} // |
<?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. ?> |
<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) {
var sum = 0;
for (i = 1; i <= n; i++)
sum += i * ((i + 1) / 2);
return sum;
}
// Driver code
var n = 4;
document.write(seriesSum(n));
// This code contributed by Rajput-Ji </script> |
20
Time Complexity: O(N), for traversing from 1 till N to calculate the required sum.
Auxiliary Space: O(1), as constant extra space is required.
An efficient solution is to use direct formula n(n+1)(n+2)/6
Mathematically, we need to find, ? ((i * (i + 1))/2), where 1 <= i <= n
So, lets solve this summation,
Sum = ? ((i * (i + 1))/2), where 1 <= i <= n = (1/2) * ? (i * (i + 1)) = (1/2) * ? (i2 + i) = (1/2) * (? i2 + ? i) We know ? i2 = n * (n + 1) * (2*n + 1) / 6 and ? i = n * ( n + 1) / 2. Substituting the value, we get, Sum = (1/2) * ((n * (n + 1) * (2*n + 1) / 6) + (n * ( n + 1) / 2)) = n * (n + 1)/2 [(2n + 1)/6 + 1/2] = n * (n + 1) * (n + 2) / 6
Below is the implementation of the above approach:
/* 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 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 |
# 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# 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 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. ?> |
<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 var n = 4;
document.write( seriesSum(n)); // This code is contributed by shikhasingrajput </script> |
20
Time Complexity: O(1), as constant operations are being performed.
Auxiliary Space: O(1), as constant extra space is required.