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.
C++
// 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; } |
Java
// 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 )); } } |
Python3
# 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 |
C#
// 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
<?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. ?> |
Javascript
<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)
Sum of the series 1.2.3 + 2.3.4 + … + n(n+1)(n+2)
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 .
C++
// 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
// 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 )); } } |
Python3
# 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 |
C#
// 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 // 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. ?> |
Javascript
<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)
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