Find the sum up to n terms of the series: 1.2.3 + 2.3.4 + … + n(n+1)(n+2). In this 1.2.3 represent the first term and 2.3.4 represent the second term .
Examples :
Input : 2
Output : 30
Explanation: 1.2.3 + 2.3.4 = 6 + 24 = 30Input : 3
Output : 90
Simple Approach We run a loop for i = 1 to n, and find the sum of (i)*(i+1)*(i+2).
And at the end display the sum .
// CPP program to find sum of the series // 1.2.3 + 2.3.4 + 3.4.5 + ... #include <bits/stdc++.h> using namespace std;
int sumofseries( int n)
{ int res = 0;
for ( int i = 1; i <= n; i++)
res += (i) * (i + 1) * (i + 2);
return res;
} // Driver Code int main()
{ cout << sumofseries(3) << endl;
return 0;
} |
// Java program to find sum of the series // 1.2.3 + 2.3.4 + 3.4.5 + ... import java.io.*;
import java.math.*;
class GFG
{ static int sumofseries( int n)
{
int res = 0 ;
for ( int i = 1 ; i <= n; i++)
res += (i) * (i + 1 ) * (i + 2 );
return res;
}
// Driver Code
public static void main(String[] args)
{
System.out.println(sumofseries( 3 ));
}
} |
# Python 3 program to find sum of the series # 1.2.3 + 2.3.4 + 3.4.5 + ... def sumofseries(n):
res = 0
for i in range ( 1 , n + 1 ):
res + = (i) * (i + 1 ) * (i + 2 )
return res
# Driver Program print (sumofseries( 3 ))
# This code is contributed # by Smitha Dinesh Semwal |
// Java program to find sum of the series // 1.2.3 + 2.3.4 + 3.4.5 + ... using System;
class GFG
{ static int sumofseries( int n)
{
int res = 0;
for ( int i = 1; i <= n; i++)
res += (i) * (i + 1) * (i + 2);
return res;
}
// Driver Code
public static void Main()
{
Console.WriteLine(sumofseries(3));
}
} // This code is contributed by vt_m. |
<?php // PHP program to find // sum of the series // 1.2.3 + 2.3.4 + 3.4.5 + ... function sumofseries( $n )
{ $res = 0;
for ( $i = 1; $i <= $n ; $i ++)
$res += ( $i ) * ( $i + 1) *
( $i + 2);
return $res ;
} // Driver Code
echo sumofseries(3);
//This code is contributed by anuj_67. ?> |
<script> // JavaScript program to find sum of the series // 1.2.3 + 2.3.4 + 3.4.5 + ... function sumofseries(n)
{
let res = 0;
for (let i = 1; i <= n; i++)
res += (i) * (i + 1) * (i + 2);
return res;
}
// Driver Code document.write(sumofseries(3));
// This code is contributed by code_hunt. </script> |
90
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach
Using Efficient Approach we know that we have to find = summation of( (n)*(n+1)*(n+2) )
Sn = summation[ (n)*(n+1)*(n+2) ]
Sn = summation [n3 + 2*n2 + n2 + 2*n]
We know sum of cubes of natural numbers is (n*(n+1))/2)2, sum of squares of natural numbers is n * (n + 1) * (2n + 1) / 6 and sum of first n natural numbers is n(n+1)/2
Sn = ((n*(n+1))/2)2 + 3((n)*(n+1)*(2*n+1)/6) + 2*((n)*(n+1)/2)
So by evaluating the above we get,
Sn = (n*(n+1)*(n+2)*(n+3)/4)
Hence it has a O(1) complexity.
// Efficient CPP program to // find sum of the series // 1.2.3 + 2.3.4 + 3.4.5 + ... #include <bits/stdc++.h> using namespace std;
// function to calculate // sum of series int sumofseries( int n)
{ return (n * (n + 1) *
(n + 2) * (n + 3) / 4);
} // Driver Code int main()
{ cout << sumofseries(3) << endl;
return 0;
} |
// Efficient Java program to // find sum of the series // 1.2.3 + 2.3.4 + 3.4.5 + .. import java.io.*;
import java.math.*;
class GFG
{ static int sumofseries( int n)
{
return (n * (n + 1 ) *
(n + 2 ) * (n + 3 ) / 4 );
}
// Driver Code
public static void main(String[] args)
{
System.out.println(sumofseries( 3 ));
}
} |
# Efficient CPP program to find sum of the # series 1.2.3 + 2.3.4 + 3.4.5 + ... # function to calculate sum of series def sumofseries(n):
return int (n * (n + 1 ) * (n + 2 ) * (n + 3 ) / 4 )
# Driver program print (sumofseries( 3 ))
# This code is contributed # by Smitha Dinesh Semwal |
// Efficient C# program to // find sum of the series // 1.2.3 + 2.3.4 + 3.4.5 + .. using System;
class GFG
{ static int sumofseries( int n)
{
return (n * (n + 1) *
(n + 2) * (n + 3) / 4);
}
// Driver Code
public static void Main()
{
Console.WriteLine(sumofseries(3));
}
} // This code is contributed by anuj_67. |
<?php // Efficient CPP program // to find sum of the // series 1.2.3 + 2.3.4 // + 3.4.5 + ... // function to calculate // sum of series function sumofseries( $n )
{ return ( $n * ( $n + 1) *
( $n + 2) * ( $n + 3) / 4);
} // Driver Code
echo sumofseries(3);
// This code is contributed by anuj_67. ?> |
<script> // Efficient Javascript program // to find sum of the // series 1.2.3 + 2.3.4 // + 3.4.5 + ... // function to calculate // sum of series function sumofseries(n)
{ return (n * (n + 1) *
(n + 2) * (n + 3) / 4);
} // Driver Code
document.write(sumofseries(3));
// This code is contributed by gfgking </script> |
90
Time Complexity: O(1)
Auxiliary Space: O(1)