Promedio de los primeros n números naturales pares

Dado un número n, encuentre el promedio de los primeros n números naturales pares 
Ej. = 2 + 4 + 6 + 8 + 10 + 12 +………+ 2n. 
Ejemplos: 
 

Input  : 7
Output : 8
(2 + 4 + 6 + 8 + 10 + 12 + 14)/7 = 8 

Input  : 5
Output : 6
(2 + 4 + 6 + 8 + 10)/5 = 6

Enfoque ingenuo: en este programa, iterar el ciclo, encontrar la suma total de los primeros n números pares y dividirlos por n. Toma 0 (N) tiempo.
 

// C++ implementation to find Average
// of sum of first n natural even numbers
#include <bits/stdc++.h>
using namespace std;
 
// function to find average of
// sum of first n even numbers
int avg_of_even_num(int n)
{
    // sum of first n even numbers
    int sum = 0;
    for (int i = 1; i <= n; i++)
        sum += 2*i;
 
    // calculating Average
    return sum/n;
}
 
// Driver Code
int main()
{
    int n = 9;
    cout << avg_of_even_num(n);
    return 0;
}

// java implementation to find Average
// of sum of first n natural even number
import java.io.*;
 
class GFG {
     
    // function to find average of
    // sum of first n even numbers
    static int avg_of_even_num(int n)
    {
     
    // sum of first n even numbers
    int sum = 0;
     
     
    for (int i = 1; i <= n; i++)
        sum += 2*i;
 
    // calculating Average
    return (sum / n);
    }
    public static void main (String[] args) {
     
    int n = 9;
    System.out.print(avg_of_even_num(n));
             
    }
}
 
// this code is contributed by 'vt_m'

# Python3 implementation to
# find Average of sum of
# first n natural even
# number
 
# Function to find average
# of sum of first n even
# numbers
def avg_of_even_num(n):
     
    # sum of first n even
    # numbers
    sum=0
    for i in range(1, n + 1):
        sum=sum + 2 * i
     
    # calculating Average
    return sum / n
 
n=9
print(avg_of_even_num(n))
 
# This code is contributed by upendra singh bartwal

// C# implementation to find
// Average of sum of first
// n natural even number
using System;
 
class GFG {
     
    // function to find average of
    // sum of first n even numbers
    static int avg_of_even_num(int n)
    {
     
    // sum of first n even numbers
    int sum = 0;
     
    for (int i = 1; i <= n; i++)
        sum += 2 * i;
 
    // calculating Average
    return (sum / n);
    }
     
    // driver code
    public static void Main () {
     
    int n = 9;
    Console.Write(avg_of_even_num(n));
             
    }
}
 
// This code is contributed by 'vt_m'

<?php
// PHP implementation to find Average
// of sum of first n natural even numbers
 
// function to find average of
// sum of first n even numbers
function avg_of_even_num($n)
{
    // sum of first n even numbers
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
        $sum += 2 * $i;
 
    // calculating Average
    return $sum / $n;
}
 
// Driver Code
$n = 9;
echo(avg_of_even_num($n));
 
// This code is contributed by Ajit.
?>

<script>
// javascript implementation to find Average
// of sum of first n natural even numbers
 
// function to find average of
// sum of first n even numbers
function avg_of_even_num( n)
{
 
    // sum of first n even numbers
    let sum = 0;
    for (let i = 1; i <= n; i++)
        sum += 2*i;
 
    // calculating Average
    return sum/n;
}
 
// Driver Code
    let n = 9;
    document.write(avg_of_even_num(n));
 
// This code is contributed by todaysgaurav
 
</script>

Producción : 
 

  10

Complejidad de tiempo: O (N) 
Método 2: – La idea es que la suma de los primeros n números pares es n (n + 1), para encontrar el promedio de los primeros n números pares divididos por n, por lo tanto, la fórmula es n (n + 1 ) / norte = ( norte + 1) . es decir, el promedio de los primeros n números pares es n+1 . tarda 0 (1) tiempo. 
 

                  Avg of sum of N even natural number = (N + 1)

Prueba 
 

Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.

Here, a = 2, d = 2, applying these values to eq.(i), get
Sum = (n/2) * [2*2 + (n-1)*2]
    = (n/2) * [4 + 2*n - 2]
    = (n/2) * (2*n + 2)
    = n * (n + 1)

 finding the Avg so divided by n = n*(n+1)/n
                                      = (n+1)

// CPP Program to find the average
// of sum of first n even numbers
#include <bits/stdc++.h>
using namespace std;
 
// Return the average of sum
// of first n even numbers
int avg_of_even_num(int n)
{
    return n+1;
}
     
// Driver Code
int main()
{
    int n = 8;
    cout << avg_of_even_num(n) << endl;
    return 0;
}

// Java Program to find the average
// of sum of first n even numbers
import java.io.*;
 
class GFG
{
 
    // Return the average of sum
    // of first n even numbers
    static int avg_of_even_num(int n)
    {
        return n + 1;
    }
     
    public static void main (String[] args) {
         
        int n = 8;
        System.out.println(avg_of_even_num(n));
         
    }
}
 
// This code is contributed by vt_m

# Python 3 Program to
# find the average
# of sum of first n
# even numbers
 
# Return the average of sum
# of first n even numbers
def avg_of_even_num(n) :
     
    return n+1
     
      
# Driven Program
n = 8
print(avg_of_even_num(n))
 
 
# This code is contributed
# by Nikita Tiwari.

// C# Program to find the average
// of sum of first n even numbers
using System;
 
class GFG {
 
    // Return the average of sum
    // of first n even numbers
    static int avg_of_even_num(int n)
    {
        return n + 1;
    }
     
    // driver code   
    public static void Main () {
         
        int n = 8;
        Console.Write(avg_of_even_num(n));
         
    }
}
 
// This code is contributed by vt_m

<?php
// PHP Program to find the average
// of sum of first n even numbers
 
// Return the average of sum
// of first n even numbers
function avg_of_even_num($n)
{
    return $n + 1;
}
     
// Driver Code
$n = 8;
echo(avg_of_even_num($n));
 
// This code is contributed by Ajit.
?>

<script>
// javascript Program to find the average
// of sum of first n even numbers
 
// Return the average of sum
// of first n even numbers
function avg_of_even_num(n)
{
    return n + 1;
}
var n = 8;
document.write(avg_of_even_num(n));
 
// This code is contributed by Amit Katiyar
</script>

Producción: 
 

9

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)