Suma del cuadrado de los primeros n números impares

Dado un número n, encuentre la suma de los cuadrados de los primeros n números naturales impares.

Ejemplos: 

Input : 3
Output : 35 
12 + 32 + 52 = 35

Input : 8
Output : 680
12 + 32 + 52 + 72 + 92 + 112 + 132 + 152 

Una solución simple es atravesar n números impares y encontrar la suma de los cuadrados. 

A continuación se muestra la implementación del enfoque.  

// Simple C++ method to find sum of square of
// first n odd numbers.
#include <iostream>
using namespace std;
 
int squareSum(int n)
{
    int sum = 0;
    for (int i = 1; i <=  n; i++)
        sum += (2*i - 1) * (2*i - 1);
    return sum;
}
 
int main()
{
    cout << squareSum(8);
    return 0;
}

// Simple Java method to
// find sum of square of
// first n odd numbers.
 
import java.io.*;
 
class GFG {
     
    static int squareSum(int n)
    {
        int sum = 0;
        for (int i = 1; i <=  n; i++)
            sum += (2*i - 1) * (2*i - 1);
        return sum;
    }
      
    //Driver Code
    public static void main(String args[])
    {  
        System.out.println(squareSum(8));
    }
}
 
// This code is contributed by
// Nikita tiwari.

# Simple Python method
# to find sum of square
# of first n odd numbers.
def squareSum(n):
     
    sm = 0
    for i in range(1, n + 1):
        sm += (2 * i - 1) * (2 * i - 1)
         
    return sm
 
# Driver Code
n=8
print(squareSum(n))
 
# This code is contributed by Ansu Kumari

// Simple C# method to find
// sum of square of first
// n odd numbers.
using System;
 
class GFG {
     
    static int squareSum(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++)
            sum += (2*i - 1) * (2*i - 1);
        return sum;
    }
     
    // Driver Code
    public static void Main()
    {
        Console.Write(squareSum(8));
    }
}
 
// This code is contributed by
// vt_m.

<?php
// Simple PHP method to find sum
// of square of first n odd numbers.
 
function squareSum( $n)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
        $sum += (2*$i - 1) * (2*$i - 1);
    return $sum;
}
 
 
    echo squareSum(8);
 
// This code is contributed by Vt_m.
?>

<script>
 
// Simple Javascript method to find
// sum of square of first n odd numbers.
function squareSum(n)
{
    let sum = 0;
    for(let i = 1; i <=  n; i++)
        sum += (2 * i - 1) * (2 * i - 1);
         
    return sum;
}
 
// Driver code
document.write(squareSum(8));
 
// This code is contributed by souravmahato348
 
</script>

Producción : 

680

Complejidad de tiempo: O(n), donde n representa el entero dado.
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.
Una solución eficiente es aplicar la siguiente fórmula.

sum = n * (4n2 - 1) / 3

How does it work? 
Please refer sum of squares of even and odd
numbers for proof.

// Efficient C++ method to find sum of
// square of first n odd numbers.
#include <iostream>
using namespace std;
 
int squareSum(int n)
{
    return n*(4*n*n - 1)/3;
}
 
int main()
{
    cout << squareSum(8);
    return 0;
}

// Efficient Java method
// to find sum of
// square of first n odd numbers.
 
import java.io.*;
 
class GFG {
     
    static int squareSum(int n)
    {
        return n*(4*n*n - 1)/3;
    }
      
    public static void main(String args[])
    {
        System.out.println(squareSum(8));
    }
}
 
// This code is contributed by
// Nikita tiwari.

# Python3 code to find sum
# of square of first n odd numbers
 
def squareSum( n ):
     
    return int(n * ( 4 * n * n - 1) / 3)
 
# driver code
ans = squareSum(8)
print (ans)
 
# This code is contributed by Saloni Gupta

// Efficient C# method to
// find sum of square of
// first n odd numbers.
using System;
 
class GFG {
     
    static int squareSum(int n)
    {
        return n * (4 * n * n - 1)/3;
    }
     
    // driver code   
    public static void Main()
    {
        Console.Write(squareSum(8));
    }
}
 
// This code is contributed by
// Vt_m.

<?php
// Efficient PHP method to find sum of
// square of first n odd numbers.
 
function squareSum($n)
{
    return $n * (4 * $n * $n - 1) / 3;
}
 
 
    echo squareSum(8);
 
// This code is contributed by Vt_m.
?>

<script>
 
// JavaScript program to find sum of
// square of first n odd numbers.
 
    function squareSum(n)
    {
        return n*(4*n*n - 1)/3;
    }
 
// Driver code
 
            document.write(squareSum(8));
 
</script>

Producción : 

680

Complejidad de tiempo: O(1), el código se ejecutará en un tiempo constante.
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.