Dado un valor n, la tarea es encontrar la suma de la serie (1*2) + (2*3) + (3*4) + ……+ n términos
Ejemplos:
Input: n = 2 Output: 8 Explanation: (1*2) + (2*3) = 2 + 6 = 8 Input: n = 3 Output: 20 Explanation: (1*2) + (2*3) + (2*4) = 2 + 6 + 12 = 20
Solución simple Uno por uno agrega elementos recursivamente.
A continuación se muestra la implementación.
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return sum
int sum(int n)
{
if (n == 1) {
return 2;
}
else {
return (n * (n + 1) + sum(n - 1));
}
}
// Driver code
int main()
{
int n = 2;
cout << sum(n);
}
Java
// Java implementation of the approach
class Solution {
// Function to return a the required result
static int sum(int n)
{
if (n == 1) {
return 2;
}
else {
return (n * (n + 1) + sum(n - 1));
}
}
// Driver code
public static void main(String args[])
{
int n = 2;
System.out.println(sum(n));
}
}
Python3
# Python3 implementation of the approach # Function to return sum def sum(n): if (n == 1): return 2; else: return (n * (n + 1) + sum(n - 1)); # Driver code n = 2; print(sum(n)); # This code is contributed by mits
C#
// Csharp implementation of the approach
using System;
class Solution {
// Function to return a the required result
static int sum(int n)
{
if (n == 1) {
return 2;
}
else {
return (n * (n + 1) + sum(n - 1));
}
}
// Driver code
public static void Main()
{
int n = 2;
Console.WrieLine(sum(n));
}
}
PHP
<?php
// PHP implementation of the approach
// Function to return sum
function sum($n)
{
if ($n == 1)
{
return 2;
}
else
{
return ($n * ($n + 1) +
sum($n - 1));
}
}
// Driver code
$n = 2;
echo sum($n);
// This code is contributed by mits
?>
Javascript
<script>
// JavaScript implementation of the approach
// Function to return sum
function sum(n)
{
if (n == 1) {
return 2;
}
else {
return (n * (n + 1) + sum(n - 1));
}
}
// Driver code
n = 2;
document.write(sum(n));
// This code is contributed by rutvik_56.
</script>
Producción:
8
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(n)
Solución Eficiente Podemos resolver este problema usando la fórmula directa.
La suma se puede escribir de la siguiente manera
: ∑(n * (n+1))
∑(n*n + n)
= ∑(n*n) + ∑(n)
Podemos aplicar las fórmulas para la suma de cuadrados de números naturales y la suma de números naturales.
= n(n+1)(2n+1)/6 + n*(n+1)/2
= n * (n + 1) * (n + 2) / 3
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return sum
int sum(int n)
{
return n * (n + 1) * (n + 2) / 3;
}
// Driver code
int main()
{
int n = 2;
cout << sum(n);
}
Java
// Java implementation of the approach
class GFG
{
// Function to return sum
static int sum(int n)
{
return n * (n + 1) * (n + 2) / 3;
}
// Driver code
public static void main(String[] args)
{
int n = 2;
System.out.println(sum(n));
}
}
// This code is contributed by Code_Mech
Python3
# Python3 implementation of the approach. # Function to return sum def Sum(n): return n * (n + 1) * (n + 2) // 3 # Driver code if __name__ == "__main__": n = 2; print(Sum(n)) # This code is contributed # by Rituraj Jain
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return sum
static int sum(int n)
{
return n * (n + 1) * (n + 2) / 3;
}
// Driver code
public static void Main(String[] args)
{
int n = 2;
Console.WriteLine(sum(n));
}
}
// This code contributed by Rajput-Ji
PHP
<?php
// PHP implementation of the approach
// Function to return sum
function sum($n)
{
return $n * ($n + 1) * ($n + 2) / 3;
}
// Driver code
$n = 2;
echo sum($n);
// This code is contributed
// by 29AjayKumar
?>
Javascript
<script>
// Javascript implementation of the approach
// Function to return sum
function sum(n)
{
return n * (n + 1) * (n + 2) / 3;
}
// Driver code
var n = 2;
document.write(sum(n));
// This code is contributed by noob2000.
</script>
Producción:
8
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)