Find the sum of N terms of the series 1, (2+3), (4+5+6), …..
Last Updated :
20 Aug, 2022
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++
#include <bits/stdc++.h>
using namespace std;
int findSum( int N)
{
return N
* (N + 1)
* (N * N + N + 2) / 8;
}
int main()
{
int N = 5;
cout << findSum(N);
}
|
Java
import java.io.*;
class GFG {
static int findSum( int N)
{
return N * (N + 1 ) * (N * N + N + 2 ) / 8 ;
}
public static void main(String[] args)
{
int N = 5 ;
System.out.println(findSum(N));
}
}
|
Python3
def findSum(N):
return N * (N + 1 ) * (N * N + N + 2 ) / / 8
if __name__ = = "__main__" :
N = 5
print (findSum(N))
|
C#
using System;
class GFG
{
static int findSum( int N)
{
return N * (N + 1) * (N * N + N + 2) / 8;
}
public static void Main()
{
int N = 5;
Console.Write(findSum(N));
}
}
|
Javascript
function findSum(N)
{
return N
* (N + 1)
* (N * N + N + 2) / 8;
}
let N = 5
document.write(findSum(N))
|
Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
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