Given a positive integer, N. Find the sum of the first N term of the series-
1, (2+3), (4+5+6),….,till N terms
Examples:
Input: N = 5
Output: 120
Input: N = 1
Output: 1
Approach: The sequence is formed by using the following pattern. For any value N-
SN = N * (N + 1) * (N2 + N + 2) / 8
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to return sum of // N term of the series int findSum( int N)
{ return N
* (N + 1)
* (N * N + N + 2) / 8;
} // Driver Code int main()
{ int N = 5;
cout << findSum(N);
} |
Java
/*package whatever //do not write package name here */ import java.io.*;
class GFG {
// Function to return sum of
// N term of the series
static int findSum( int N)
{
return N * (N + 1 ) * (N * N + N + 2 ) / 8 ;
}
// Driver Code
public static void main(String[] args)
{
int N = 5 ;
System.out.println(findSum(N));
}
} // This code is contributed by Potta Lokesh |
Python3
# Python 3 program for the above approach # Function to return sum of # N term of the series def findSum(N):
return N * (N + 1 ) * (N * N + N + 2 ) / / 8
# Driver Code if __name__ = = "__main__" :
# Value of N
N = 5
print (findSum(N))
# This code is contributed by Abhishek Thakur. |
C#
/*package whatever //do not write package name here */ using System;
class GFG
{ // Function to return sum of
// N term of the series
static int findSum( int N)
{
return N * (N + 1) * (N * N + N + 2) / 8;
}
// Driver Code
public static void Main()
{
int N = 5;
Console.Write(findSum(N));
}
} // This code is contributed by Saurabh Jaiswal |
Javascript
// Javascript program to implement // the above approach // Function to return sum of // N term of the series function findSum(N)
{ return N
* (N + 1)
* (N * N + N + 2) / 8;
} // Driver Code let N = 5 document.write(findSum(N)) // This code is contributed by saurabh_jaiswal. |
Output
120
Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.