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++ 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);
}
/*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
# 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.
/*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 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.