Dado un número n, encuentre la suma de los primeros números naturales.
C++
// CPP program to find sum of first
// n natural numbers.
#include<iostream>
using namespace std;
// Returns sum of first n natural
// numbers
int findSum(int n)
{
int sum = 0;
for (int x=1; x<=n; x++)
sum = sum + x;
return sum;
}
// Driver code
int main()
{
int n = 5;
cout << findSum(n);
return 0;
}
Java
// JAVA program to find sum of first
// n natural numbers.
import java.io.*;
class GFG{
// Returns sum of first n natural
// numbers
static int findSum(int n)
{
int sum = 0;
for (int x = 1; x <= n; x++)
sum = sum + x;
return sum;
}
// Driver code
public static void main(String args[])
{
int n = 5;
System.out.println(findSum(n));
}
}
// This code is contributed by Nikita Tiwari.
Python
# PYTHON program to find sum of first # n natural numbers. # Returns sum of first n natural # numbers def findSum(n) : sum = 0 x = 1 while x <=n : sum = sum + x x = x + 1 return sum # Driver code n = 5 print findSum(n) # This code is contributed by Nikita Tiwari.
C#
// C# program to find sum of first
// n natural numbers.
using System;
class GFG{
// Returns sum of first n natural
// numbers
static int findSum(int n)
{
int sum = 0;
for (int x = 1; x <= n; x++)
sum = sum + x;
return sum;
}
// Driver code
public static void Main()
{
int n = 5;
Console.Write(findSum(n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to find sum of first
// n natural numbers.
// Returns sum of first n natural
// numbers
function findSum($n)
{
$sum = 0;
for ($x = 1; $x <= $n; $x++)
$sum = $sum + $x;
return $sum;
}
// Driver code
$n = 5;
echo findSum($n);
// This code is contributed by Sam007
?>
Javascript
<script>
// Javascript program to find sum of first
// n natural numbers.
// Returns sum of first n natural
// numbers
function findSum(n)
{
let sum = 0;
for (let x = 1; x <= n; x++)
sum = sum + x;
return sum;
}
// Driver code
let n = 5;
document.write(findSum(n));
// This code is contributed by rishavmahato348.
</script>
C++
// Efficient CPP program to find sum of first
// n natural numbers.
#include<iostream>
using namespace std;
// Returns sum of first n natural
// numbers
int findSum(int n)
{
return n * (n + 1) / 2;
}
// Driver code
int main()
{
int n = 5;
cout << findSum(n);
return 0;
}
Java
// Efficient JAVA program to find sum
// of first n natural numbers.
import java.io.*;
class GFG{
// Returns sum of first n natural
// numbers
static int findSum(int n)
{
return n * (n + 1) / 2;
}
// Driver code
public static void main(String args[])
{
int n = 5;
System.out.println(findSum(n));
}
}
// This code is contributed by Nikita Tiwari.
Python
# Efficient CPP program to find sum # of first n natural numbers. # Returns sum of first n natural # numbers def findSum(n) : return n * (n + 1) / 2 # Driver code n = 5 print findSum(n) # This code is contributed by Nikita Tiwari.
C#
// Efficient C# program to find sum
// of first n natural numbers.
using System;
class GFG{
// Returns sum of first n natural
// numbers
static int findSum(int n)
{
return n * (n + 1) / 2;
}
// Driver code
public static void Main()
{
int n = 5;
Console.Write(findSum(n));
}
}
// This code is contributed by vt_m.
php
<?php
// Efficient PHP program to find sum
// of first n natural numbers.
// Returns sum of first n natural
// numbers
function findSum($n)
{
return ($n * ($n + 1) / 2);
}
// Driver code
$n = 5;
echo findSum($n);
// This code is contributed by Sam007
?>
Javascript
<script>
// javascript Program to find the average
// of sum of first n natural numbers
// Return the average of sum
// of first n even numbers
function findSum(n)
{
return n * (n + 1) / 2;
}
var n = 5;
document.write(findSum(n));
// This code is contributed by sravan kumar
</script>
C++
// Efficient CPP program to find sum of first
// n natural numbers that avoids overflow if
// result is going to be within limits.
#include<iostream>
using namespace std;
// Returns sum of first n natural
// numbers
int findSum(int n)
{
if (n % 2 == 0)
return (n/2) * 1LL * (n+1); // Here multiplying by 1LL help to perform calculations in long long, so that answer should not be overflowed
// If n is odd, (n+1) must be even
else
return ((n + 1) / 2) * 1LL * n; // Here multiplying by 1LL help to perform calculations in long long, so that answer should not be overflowed
}
// Driver code
int main()
{
int n = 5;
cout << findSum(n);
return 0;
}
Java
// Efficient JAVA program to find sum of first
// n natural numbers that avoids overflow if
// result is going to be within limits.
import java.io.*;
class GFG{
// Returns sum of first n natural
// numbers
static int findSum(int n)
{
if (n % 2 == 0)
return (n / 2) * (n + 1);
// If n is odd, (n+1) must be even
else
return ((n + 1) / 2) * n;
}
// Driver code
public static void main(String args[])
{
int n = 5;
System.out.println(findSum(n));
}
}
//This code is contributed by Nikita Tiwari.
Python
# Efficient Python program to find the sum # of first n natural numbers that avoid # overflow if the result is going to be # within limits. # Returns sum of first n natural # numbers def findSum(n) : if (n % 2 == 0) : return (n / 2) * (n + 1) # If n is odd, (n+1) must be even else : return ((n + 1) / 2) * n # Driver code n = 5 print findSum(n) # This code is contributed by Nikita Tiwari.
C#
// Efficient C# program to find the sum of first
// n natural numbers that avoid overflow if
// result is going to be within limits.
using System;
class GFG{
// Returns sum of first n natural
// numbers
static int findSum(int n)
{
if (n % 2 == 0)
return (n / 2) * (n + 1);
// If n is odd, (n+1) must be even
else
return ((n + 1) / 2) * n;
}
// Driver code
public static void Main()
{
int n = 5;
Console.Write(findSum(n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// Efficient php program to find sum of first
// n natural numbers that avoids overflow if
// result is going to be within limits.
// Returns sum of first n natural
// numbers
function findSum($n)
{
if ($n % 2 == 0)
return ($n / 2) *
($n + 1);
// If n is odd, (n+1) must be even
else
return (($n + 1) / 2) * $n;
}
// Driver code
$n = 5;
echo findSum($n);
// This code is contributed by Sam007
?>
Javascript
<script>
//efficient approach using javascript to find the average
// of sum of first n natural numbers
// Return the average of sum
// of first n even numbers
function findSum(n)
{
if (n % 2 == 0)
return (n / 2) * (n + 1)
// If n is odd, (n+1) must be even
else
return ((n + 1) / 2) * n
}
var n = 5;
document.write(findSum(n));
// This code is contributed by sravan kumar
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA