Imprime la suma de la serie 1 3 + 2 3 + 3 3 + 4 3 + …….+ n 3 hasta el n-ésimo término.
Ejemplos:
Input : n = 5 Output : 225 13 + 23 + 33 + 43 + 53 = 225 Input : n = 7 Output : 784 13 + 23 + 33 + 43 + 53 + 63 + 73 = 784
Una solución simple es agregar términos uno por uno.
C++
// Simple C++ program to find sum of series // with cubes of first n natural numbers #include <iostream> using namespace std; /* Returns the sum of series */ int sumOfSeries(int n) { int sum = 0; for (int x = 1; x <= n; x++) sum += x * x * x; return sum; } // Driver Function int main() { int n = 5; cout << sumOfSeries(n); return 0; }
Java
// Simple Java program to find sum of series // with cubes of first n natural numbers import java.util.*; import java.lang.*; class GFG { /* Returns the sum of series */ public static int sumOfSeries(int n) { int sum = 0; for (int x = 1; x <= n; x++) sum += x * x * x; return sum; } // Driver Function public static void main(String[] args) { int n = 5; System.out.println(sumOfSeries(n)); } } // Code Contributed by Mohit Gupta_OMG <(0_o)>
Python3
# Simple Python program to find sum of series # with cubes of first n natural numbers # Returns the sum of series def sumOfSeries(n): sum = 0 for i in range(1, n + 1): sum += i * i*i return sum # Driver Function n = 5 print(sumOfSeries(n)) # Code Contributed by Mohit Gupta_OMG <(0_o)>
C#
// Simple C# program to find sum of series // with cubes of first n natural numbers using System; class GFG { /* Returns the sum of series */ static int sumOfSeries(int n) { int sum = 0; for (int x = 1; x <= n; x++) sum += x * x * x; return sum; } // Driver Function public static void Main() { int n = 5; Console.Write(sumOfSeries(n)); } } // This code is contributed by // Smitha Dinesh Semwal
PHP
<?php // Simple PHP program to find sum of series // with cubes of first n natural numbers // Returns the sum of series function sumOfSeries( $n) { $sum = 0; for ($x = 1; $x <= $n; $x++) $sum += $x * $x * $x; return $sum; } // Driver code $n = 5; echo sumOfSeries($n); // This Code is contributed by vt_m. ?>
Javascript
<script> // Simple javascript program to find sum of series // with cubes of first n natural numbers /* Returns the sum of series */ function sumOfSeries( n) { let sum = 0; for (let x = 1; x <= n; x++) sum += x * x * x; return sum; } // Driven Program let n = 5; document.write(sumOfSeries(n)); // This code contributed by aashish1995 </script>
Producción :
225
Complejidad de tiempo: O(n)
Espacio auxiliar: O(1)
Una solución eficiente es utilizar la fórmula matemática directa que es (n ( n + 1 ) / 2) ^ 2
For n = 5 sum by formula is (5*(5 + 1 ) / 2)) ^ 2 = (5*6/2) ^ 2 = (15) ^ 2 = 225 For n = 7, sum by formula is (7*(7 + 1 ) / 2)) ^ 2 = (7*8/2) ^ 2 = (28) ^ 2 = 784
C++
// A formula based C++ program to find sum // of series with cubes of first n natural // numbers #include <iostream> using namespace std; int sumOfSeries(int n) { int x = (n * (n + 1) / 2); return x * x; } // Driver Function int main() { int n = 5; cout << sumOfSeries(n); return 0; }
Java
// A formula based Java program to find sum // of series with cubes of first n natural // numbers import java.util.*; import java.lang.*; class GFG { /* Returns the sum of series */ public static int sumOfSeries(int n) { int x = (n * (n + 1) / 2); return x * x; } // Driver Function public static void main(String[] args) { int n = 5; System.out.println(sumOfSeries(n)); } } // Code Contributed by Mohit Gupta_OMG <(0_o)>
Python3
# A formula based Python program to find sum # of series with cubes of first n natural # numbers # Returns the sum of series def sumOfSeries(n): x = (n * (n + 1) / 2) return (int)(x * x) # Driver Function n = 5 print(sumOfSeries(n)) # Code Contributed by Mohit Gupta_OMG <(0_o)>
C#
// A formula based C# program to // find sum of series with cubes // of first n natural numbers using System; class GFG { // Returns the sum of series public static int sumOfSeries(int n) { int x = (n * (n + 1) / 2); return x * x; } // Driver Function public static void Main() { int n = 5; Console.Write(sumOfSeries(n)); } } // Code Contributed by nitin mittal.
PHP
<?php // A formula based PHP program to find sum // of series with cubes of first n natural // numbers function sumOfSeries($n) { $x = ($n * ($n + 1) / 2); return $x * $x; } // Driver Function $n = 5; echo sumOfSeries($n); // This code is contributed by vt_m. ?>
Javascript
<script> // Simple javascript program to find sum of series // with cubes of first n natural numbers /* Returns the sum of series */ function sumOfSeries( n) { x = (n * (n + 1) / 2) return (x * x) } // Driven Program let n = 5; document.write(sumOfSeries(n)); // This code is contributed by sravan kumar </script>
Producción:
225
Complejidad de tiempo: O(1)
Espacio auxiliar: O(1)
¿Cómo funciona esta fórmula?
Podemos probar la fórmula usando inducción matemática. Podemos ver fácilmente que la fórmula se cumple para n = 1 y n = 2. Sea esto cierto para n = k-1.
Let the formula be true for n = k-1. Sum of first (k-1) natural numbers = [((k - 1) * k)/2]2 Sum of first k natural numbers = = Sum of (k-1) numbers + k3 = [((k - 1) * k)/2]2 + k3 = [k2(k2 - 2k + 1) + 4k3]/4 = [k4 + 2k3 + k2]/4 = k2(k2 + 2k + 1)/4 = [k*(k+1)/2]2
El programa anterior provoca un desbordamiento, incluso si el resultado no supera el límite de números enteros. Al igual que en la publicación anterior , podemos evitar el desbordamiento hasta cierto punto haciendo primero la división.
C++
// Efficient CPP program to find sum of cubes // of first n natural numbers that avoids // overflow if result is going to be with in // limits. #include <iostream> using namespace std; // Returns sum of first n natural // numbers int sumOfSeries(int n) { int x; if (n % 2 == 0) x = (n / 2) * (n + 1); else x = ((n + 1) / 2) * n; return x * x; } // Driver code int main() { int n = 5; cout << sumOfSeries(n); return 0; }
Java
// Efficient Java program to find sum of cubes // of first n natural numbers that avoids // overflow if result is going to be with in // limits. import java.util.*; import java.lang.*; class GFG { /* Returns the sum of series */ public static int sumOfSeries(int n) { int x; if (n % 2 == 0) x = (n / 2) * (n + 1); else x = ((n + 1) / 2) * n; return x * x; } // Driver Function public static void main(String[] args) { int n = 5; System.out.println(sumOfSeries(n)); } } // Code Contributed by Mohit Gupta_OMG <(0_o)>
Python3
# Efficient Python program to find sum of cubes # of first n natural numbers that avoids # overflow if result is going to be with in # limits. # Returns the sum of series def sumOfSeries(n): x = 0 if n % 2 == 0 : x = (n / 2) * (n + 1) else: x = ((n + 1) / 2) * n return (int)(x * x) # Driver Function n = 5 print(sumOfSeries(n)) # Code Contributed by Mohit Gupta_OMG <(0_o)>
C#
// Efficient C# program to find sum of // cubes of first n natural numbers // that avoids overflow if result is // going to be with in limits. using System; class GFG { /* Returns the sum of series */ public static int sumOfSeries(int n) { int x; if (n % 2 == 0) x = (n / 2) * (n + 1); else x = ((n + 1) / 2) * n; return x * x; } // Driver code static public void Main () { int n = 5; Console.WriteLine(sumOfSeries(n)); } } // This code is contributed by Ajit.
PHP
<?php // Efficient PHP program to // find sum of cubes of first // n natural numbers that avoids // overflow if result is going // to be with in limits. // Returns sum of first n // natural numbers function sumOfSeries($n) { $x; if ($n % 2 == 0) $x = ($n / 2) * ($n + 1); else $x = (($n + 1) / 2) * $n; return $x * $x; } // Driver code $n = 5; echo sumOfSeries($n); // This code is contributed by vt_m. ?>
Javascript
<script> // Simple javascript program to find sum of series // with cubes of first n natural numbers /* Returns the sum of series */ function sumOfSeries( n) { x=0 if (n % 2 == 0) x = (n / 2) * (n + 1) else x = ((n + 1) / 2) * n return (x * x) } // Driven Program let n = 5; document.write(sumOfSeries(n)); // This code contributed by sravan </script>
Producción:
225
Complejidad temporal : O(1) desde la realización de operaciones constantes
Espacio auxiliar: O(1)
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA