Encuentra la suma de los primeros N términos de la serie 2*3*5, 3*5*7, 4*7*9, …

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

(2*3*5), (3*5*7), (4*7*9),… 
 

Ejemplos: 
 

Entrada: N = 3 
Salida: 387 
S ₃ = (2 * 3 * 5) + (3 * 5 * 7) + (4 * 7 * 9) = 30 + 105 + 252 = 387
Entrada: N = 5 
Salida: 1740 
 

Método: Sea T ^{n el} _término N de la serie . La suma de la serie se puede encontrar fácilmente observando el ^término N de la serie: 
 

T _n = {n ^ésimo término de 2, 3, 4, …} * {n ^ésimo término de 3, 5, 7, …} * {n ^ésimo término de 5, 7, 9, …} 
T _n = (n + 1) * (2 * n + 1) * (2* n + 3) 
T _n = 4n ³ + 12n ² + 11n + 3 
 

La suma ( S _n ) de los primeros n términos se puede encontrar mediante
 

S norte = Σ T _{norte S  norte} = Σ[4n ³ + 12n ² + 11n + 3] 
S norte _{= ( n} / 2) * [2n ³ + 12n ² + 25n + 21]
 
 

A continuación se muestra la implementación del enfoque anterior:
 

// C++ program to find sum of the
// first n terms of the given series
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the sum of the
// first n terms of the given series
int calSum(int n)
{
    // As described in the approach
    return (n * (2 * n * n * n + 12 * n * n + 25 * n + 21)) / 2;
}
 
// Driver code
int main()
{
    int n = 3;
    cout << calSum(n);
    return 0;
}

// Java program to find sum of the
// first n terms of the given series
class GFG {
 
    // Function to return the sum of the
    // first n terms of the given series
    static int calSum(int n)
    {
 
        // As described in the approach
        return (n * (2 * n * n * n + 12 * n * n + 25 * n + 21)) / 2;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int n = 3;
        System.out.println(calSum(n));
    }
}

# C++ program to find sum of the
# first n terms of the given series
 
# Function to return the sum of the
# first n terms of the given series
def calSum(n):
     
    # As described in the approach
    return (n*(2 * n*n * n + 12 * n*n + 25 * n + 21))/2;
 
# Driver Code
n = 3
print(calSum(n))

// C# program to find sum of the
// first n terms of the given series
using System;
 
class GFG {
 
    // Function to return the sum of the
    // first n terms of the given series
    static int calSum(int n)
    {
 
        // As described in the approach
        return (n * (2 * n * n * n + 12 * n * n + 25 * n + 21)) / 2;
    }
 
    // Driver code
    static public void Main()
    {
        int n = 3;
        Console.WriteLine(calSum(n));
    }
}

<?php
// PHP script to find sum of the
// first n terms of the given series
 
// Function to return the sum of the
// first n terms of the given series
function calculateSum($n)
{
     
    // As described in the approach
    return ($n*(2*$n*$n*$n+12*$n*$n+25*$n+21))/2;
}
 
// Driver code
$n = 3;
echo calculateSum($n);
 
?>

<script>
// Javascript program to find sum of the
// first n terms of the given series
 
    // Function to return the sum of the
    // first n terms of the given series
    function calSum( n) {
 
        // As described in the approach
        return (n * (2 * n * n * n + 12 * n * n + 25 * n + 21)) / 2;
    }
 
    // Driver Code
    let n = 3;
    document.write(calSum(n));
 
// This code is contributed by 29AjayKumar
</script>

387

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)