Dado un entero positivo n y la tarea es encontrar la suma de la serie 1*2*3 + 2*3*4 + 4*5*6 + . . .+ n*(n+1)*(n+2).
Ejemplos:
Input : n = 10 Output : 4290 1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + 5*6*7 + 6*7*8 + 7*8*9 + 8*9*10 + 9*10*11 + 10*11*12 = 6 + 24 + 60 + 120 + 210 + 336 + 504 + 720 + 990 + 1320 = 4290 Input : n = 7 Output : 1260
Método 1: en este caso, el bucle se ejecutará n veces y calculará la suma.
C++
// Program to find the sum of series
// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)
#include <bits/stdc++.h>
using namespace std;
// Function to calculate sum of series.
int sumOfSeries(int n)
{
int sum = 0;
for (int i = 1; i <= n; i++)
sum = sum + i * (i + 1) * (i + 2);
return sum;
}
// Driver function
int main()
{
int n = 10;
cout << sumOfSeries(n);
return 0;
}
Java
// Java Program to find the sum of series
// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)
public class GfG{
// Function to calculate sum of series.
static int sumOfSeries(int n)
{
int sum = 0;
for (int i = 1; i <= n; i++)
sum = sum + i * (i + 1) * (i + 2);
return sum;
}
// Driver Code
public static void main(String s[])
{
int n = 10;
System.out.println(sumOfSeries(n));
}
}
// This article is contributed by Gitanjali.
Python3
# Python program to find the # sum of series # 1*2*3 + 2*3*4 + . . . # + n*(n+1)*(n+1) # Function to calculate sum # of series. def sumOfSeries(n): sum = 0; i = 1; while i<=n: sum = sum + i * (i + 1) * ( i + 2) i = i + 1 return sum # Driver code n = 10 print(sumOfSeries(n)) # This code is contributed by "Abhishek Sharma 44"
C#
// C# Program to find the sum of series
// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)
using System;
public class GfG
{
// Function to calculate sum of series.
static int sumOfSeries(int n)
{
int sum = 0;
for (int i = 1; i <= n; i++)
sum = sum + i * (i + 1) * (i + 2);
return sum;
}
// Driver Code
public static void Main()
{
int n = 10;
Console.WriteLine(sumOfSeries(n));
}
}
// This article is contributed by vt_m.
PHP
<?php
// PHP Program to find the sum of series
// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)
// Function to calculate sum of series.
function sumOfSeries($n)
{
$sum = 0;
for ($i = 1; $i <= $n; $i++)
$sum = $sum + $i * ($i + 1) *
($i + 2);
return $sum;
}
// Driver Code
$n = 10;
echo sumOfSeries($n);
// This code is contributed by vt_m.
?>
Javascript
<script>
// Javascript Program to find the sum of series
// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)
// Function to calculate sum of series.
function sumOfSeries( n) {
let sum = 0;
for ( let i = 1; i <= n; i++)
sum = sum + i * (i + 1) * (i + 2);
return sum;
}
// Driver Code
let n = 10;
document.write(sumOfSeries(n));
// This code contributed by Princi Singh
</script>
Producción:
4290
Complejidad de tiempo: O(n)
Espacio auxiliar: O(1)
Método 2: En este caso usamos la fórmula para sumar la suma de las series.
Given series 1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + . . . + n*(n+1)*(n+2)
sum of series = (n * (n+1) * (n+2) * (n+3)) / 4
Put n = 10 then
sum = (10 * (10+1) * (10+2) * (10+3)) / 4
= (10 * 11 * 12 * 13) / 4
= 4290
C++
// Program to find the sum of series
// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)
#include <bits/stdc++.h>
using namespace std;
// Function to calculate sum of series.
int sumOfSeries(int n)
{
return (n * (n + 1) * (n + 2) * (n + 3)) / 4;
}
// Driver function
int main()
{
int n = 10;
cout << sumOfSeries(n);
return 0;
}
Java
// Program to find the
// sum of series
// 1*2*3 + 2*3*4 +
// . . . + n*(n+1)*(n+1)
import java.io.*;
class GFG {
// Function to calculate
// sum of series.
static int sumOfSeries(int n)
{
return (n * (n + 1) *
(n + 2) * (n + 3)) / 4;
}
// Driver function
public static void main (String[] args) {
int n = 10;
System.out.println(sumOfSeries(n));
}
}
// This code is contributed by Nikita Tiwari.
Python3
# Python program to find the # sum of series # 1*2*3 + 2*3*4 + . . . # + n*(n+1)*(n+1) # Function to calculate sum # of series. def sumOfSeries(n): return (n * (n + 1) * (n + 2 ) * (n + 3)) / 4 #Driver code n = 10 print(sumOfSeries(n)) # This code is contributed by "Abhishek Sharma 44"
C#
// Program to find the
// sum of series
// 1*2*3 + 2*3*4 +
// . . . + n*(n+1)*(n+1)
using System;
class GFG {
// Function to calculate
// sum of series.
static int sumOfSeries(int n)
{
return (n * (n + 1) *
(n + 2) * (n + 3)) / 4;
}
// Driver function
public static void Main () {
int n = 10;
Console.WriteLine(sumOfSeries(n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP Program to find the sum of series
// 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1)
// Function to calculate sum of series.
function sumOfSeries($n)
{
return ($n * ($n + 1) * ($n + 2) *
($n + 3)) / 4;
}
// Driver Code
$n = 10;
echo sumOfSeries($n);
// This code is contributed by vt_m.
?>
Javascript
<script>
// Program to find the
// sum of series
// 1*2*3 + 2*3*4 +
// . . . + n*(n+1)*(n+1)
// Function to calculate
// sum of series.
function sumOfSeries(n) {
return (n * (n + 1) * (n + 2) * (n + 3)) / 4;
}
// Driver function
var n = 10;
document.write(sumOfSeries(n));
// This code is contributed by Amit Katiyar
</script>
Producción:
4290
Complejidad de tiempo: O(1)
Espacio Auxiliar: O(1)
¿Cómo funciona esta fórmula?
We can prove working of this formula using
mathematical induction.
According to formula, sum of (k -1) terms is
((k - 1) * (k) * (k + 1) * (k + 2)) / 4
Sum of k terms
= sum of k-1 terms + value of k-th term
= ((k - 1) * (k) * (k + 1) * (k + 2)) / 4 +
k * (k + 1) * (k + 2)
Taking common term (k + 1) * (k + 2) out.
= (k + 1)*(k + 2) [k*(k-1)/4 + k]
= (k + 1)*(k + 2) * k * (k + 3)/4
= k * (k + 1) * (k + 2) * (k + 3)/4
Publicación traducida automáticamente
Artículo escrito por Dharmendra_Kumar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA