Encuentra la suma de los primeros N términos de la serie 2×3 + 4×4 + 6×5 + 8×6 + …

Dado un número entero N . La tarea es encontrar la suma de N términos de la serie dada:
 

2×3 + 4×4 + 6×5 + 8×6 + … + hasta n términos 
 

Ejemplos: 
 

Input : N = 5
Output : Sum = 170

Input : N = 10
Output : Sum = 990

Sea t _N el término N-ésimo de la serie . t ₁ = 2 × 3 = (2 × 1)(1 + 2)  t ₂ = 4 × 4 = (2 × 2)(2 + 2)  t ₃ = 6 × 5 = (2 × 3)(3 + 2 )  t ₄ = 8 × 6 = (2 × 4)(4 + 2)  . . . t _N = (2 × N)(N + 2) La suma de n términos de la serie, 

 Sn = t1 + t2 +... + tn
    =======

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
#include<iostream>
using namespace std;
 
// calculate sum upto N term of series
void Sum_upto_nth_Term(int n)
{
    int r = n * (n + 1) *
                (2 * n + 7) / 3;
    cout << r;
}
 
// Driver code
int main()
{
    int N = 5;
    Sum_upto_nth_Term(N) ;
    return 0;
}

// Java program to find sum upto
// N term of the series:
 
import java.io.*;
 
class GFG {
// calculate sum upto N term of series
static void Sum_upto_nth_Term(int n)
{
    int r = n * (n + 1) *
                (2 * n + 7) / 3;
    System.out.println(r);
}
 
// Driver code
    public static void main (String[] args) {
    int N = 5;
    Sum_upto_nth_Term(N);
    }
}

# Python program to find sum upto N term of the series:
# 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
 
# calculate sum upto N term of series
def Sum_upto_nth_Term(n):
    return n * (n + 1) * (2 * n + 7) // 3
 
# Driver code
N = 5
print(Sum_upto_nth_Term(N))

// C# program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
 
using System;
 
class GFG
{
// calculate sum upto N term of series
static void Sum_upto_nth_Term(int n)
{
    int r = n * (n + 1) *
                (2 * n + 7) / 3;
    Console.Write(r);
}
 
// Driver code
public static void Main()
{
    int N = 5;
    Sum_upto_nth_Term(N);
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

<?php
// PHP program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
function Sum_upto_nth_Term($n)
{
    $r = $n * ($n + 1) *
              (2 * $n + 7) / 3;
    echo $r;
}
 
// Driver code
$N = 5;
Sum_upto_nth_Term($N);
 
// This code is contributed
// by Shashank_Sharma
?>

<script>
 
// Javascript program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
 
// calculate sum upto N term of series
function Sum_upto_nth_Term(n)
{
    let r = n * (n + 1) *
                (2 * n + 7) / 3;
    document.write(r);
}
 
// Driver code
 
    let N = 5;
    Sum_upto_nth_Term(N) ;
 
 
// This code is contributed by Mayank Tyagi
 
</script>

170

Complejidad Temporal: O(1), es una constante.
Espacio Auxiliar: O(1), no se requiere espacio extra.