Suma de los primeros N términos de la serie 2, 6, 12, 20, 30….

Dado un número N, la tarea es encontrar la suma de los primeros N términos de la siguiente serie:

Sn = 2 + 6 + 12 + 20 + 30 … hasta n términos

Ejemplos: 

Input: N = 2
Output: 8
Explanation: 2 + 6 = 8

Input: N = 4 
Output: 40
Explanation: 2 + 6+ 12 + 20 = 40

Enfoque: Sea, el término n-ésimo sea denotado por Sn. 
Este problema se puede resolver fácilmente dividiendo cada término de la siguiente manera: 

Sn = 2 + 6 + 12 + 20 + 30......
Sn = (1+1^2) + (2+2^2) + (3+3^2) + (4+4^2) +......
Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^2 + 2^2 + 3^2 + 4^2 + ...upto n terms)

Observamos que Sn puede descomponerse en la suma de dos series. 
Por lo tanto, la suma de los primeros n términos se da de la siguiente manera: 

Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^2 + 2^2 + 3^2 + 4^2 + ...upto n terms)
Sn = n*(n + 1)/2 + n*(n + 1)*(2*n + 1)/6

A continuación se muestra la implementación del enfoque anterior: 

// 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 + n * (n + 1) * (2 * n + 1) / 6;
}
 
// Driver code
int main()
{
    // number of terms to be
    // included in the sum
    int n = 3;
 
    // find the Sn
    cout << "Sum = " << calculateSum(n);
 
    return 0;
}

// Java program to find sum of first n terms
 
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
     
// Function to calculate the sum
static int calculateSum(int n)
{
   
    return n * (n + 1) / 2 + n * 
            (n + 1) * (2 * n + 1) / 6;
}
   
// Driver code
public static void main(String args[])
{
    // number of terms to be
    // included in the sum
    int n = 3;
   
    // find the Sn
    System.out.print("Sum = " + calculateSum(n));
   
}
}

# Python program to find sum of
# first n terms
 
# Function to calculate the sum
def calculateSum(n):
    return (n * (n + 1) // 2 + n *
           (n + 1) * (2 * n + 1) // 6)
 
# 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

// C# program to find sum of
// first n terms
using System;
 
class GFG
{
     
// Function to calculate the sum
static int calculateSum(int n)
{
 
    return n * (n + 1) / 2 + n *
               (n + 1) * (2 * n + 1) / 6;
}
 
// Driver code
public static void Main()
{
    // number of terms to be
    // included in the sum
    int n = 3;
 
    // find the Sn
    Console.WriteLine("Sum = " +
                       calculateSum(n));
}
}
 
// This code is contributed by inder_verma

<?php
// PHP program to find sum
// of first n terms
 
// Function to calculate the sum
function calculateSum($n)
{
    return $n * ($n + 1) / 2 + $n *
                ($n + 1) * (2 * $n + 1) / 6;
}
 
// Driver code
 
// number of terms to be
// included in the sum
$n = 3;
 
// find the Sn
echo "Sum = " , calculateSum($n);
 
// This code is contributed
// by inder_verma
?>

<script>
 
// Javascript program to find sum of first n terms
 
// Function to calculate the sum
function calculateSum(n)
{
 
    return n * (n + 1) / 2 + n *
            (n + 1) * (2 * n + 1) / 6;
}
 
// Driver code
 
    // number of terms to be
    // included in the sum
    let n = 3;
 
    // find the Sn
    document.write("Sum = " + calculateSum(n));
 
// This code is contributed by Mayank Tyagi
 
</script>

Sum = 20

Complejidad de tiempo: O(1), estamos usando solo operaciones de tiempo constante.
Espacio Auxiliar: O(1)