Sum of the first N terms of the series 2,10, 30, 68,….

Given a number N, the task is to find the sum of first N terms of the below series:

Sn = 2 + 10 + 30 + 68 + … upto n terms

Examples:



Input: N = 2
Output: 12
2 + 10
= 12

Input: N = 4 
Output: 40
2 + 10 + 30 + 68
= 110

Approach: Let, the nth term be denoted by tn.
This problem can easily be solved by spliting each term as follows :

Sn = 2 + 10 + 30 + 68 + ......
Sn = (1+1^3) + (2+2^3) + (3+3^3) + (4+4^3) +......
Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^3 + 2^3 + 3^3 + 4^3 + ...upto n terms)

We observed that Sn can broken down into summation of two series.
Hence, the sum of first n terms is given as follows:

Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^3 + 2^3 + 3^3 + 4^3 + ...upto n terms)
Sn = n*(n + 1)/2 + (n*(n + 1)/2)^2

Below is the implementation of above approach:

C++

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// C++ program to find sum of first n terms
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate the sum
int calculateSum(int n)
{
  
    return n * (n + 1) / 2 
           + pow((n * (n + 1) / 2), 2);
}
  
// Driver code
int main()
{
    // number of terms to be
    // included in the sum
    int n = 3;
  
    // find the Sum
    cout << "Sum = " << calculateSum(n);
  
    return 0;
}

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Java

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// Java program to find sum of first n terms 
  
public class GFG {
      
    // Function to calculate the sum 
    static int calculateSum(int n) 
    
        
        return n * (n + 1) / 2  
               + (int)Math.pow((n * (n + 1) / 2), 2); 
    
      
    // Driver code
    public static void main(String args[])
    {
        // number of terms to be 
        // included in the sum 
        int n = 3
        
        // find the Sum 
        System.out.println("Sum = "+ calculateSum(n)); 
    }
    // This Code is contributed by ANKITRAI1
}

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Python3

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# Python program to find sum 
# of first n terms
  
# Function to calculate the sum
def calculateSum(n):
    return (n * (n + 1) // 2 + 
        pow((n * (n + 1) // 2), 2))
  
# Driver code
  
# number of terms to be
# included in the sum
n = 3
  
# find the Sum
print("Sum = ", calculateSum(n))
  
# This code is contributed by
# Sanjit_Prasad

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C#

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// C# program to find sum of first n terms
using System;
class gfg
{
    // Function to calculate the sum
    public void calculateSum(int n)
    {
        double r = (n * (n + 1) / 2 + 
                Math.Pow((n * (n + 1) / 2), 2));
        Console.WriteLine("Sum = " + r);
    }
  
    // Driver code
    public static int Main()
    {
        gfg g = new gfg();
  
        // number of terms to be
        // included in the sum
        int n = 3;
          
        // find the Sum
        g.calculateSum(n);
        Console.Read();
        return 0;
    }
}

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PHP

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<?php
// PHP program to find sum
// of first n terms
  
// Function to calculate the sum
function calculateSum($n)
{
    return $n * ($n + 1) / 2 + 
      pow(($n * ($n + 1) / 2), 2);
}
  
// Driver code
  
// number of terms to be
// included in the sum
$n = 3;
  
// find the Sum
echo "Sum = " , calculateSum($n);
  
// This code is contributed 
// by anuj_67
?>

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Output:

Sum = 42


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