Dado un entero positivo N , la tarea es encontrar la suma hasta el N- ésimo término de la serie:
3 3 – 2 3 , 5 3 – 4 3 , 7 3 – 6 3 , …., hasta N términos
Ejemplos :
Entrada : N = 10
Salida : 4960Entrada : N = 1
Salida : 19
Enfoque ingenuo :
- Inicialice dos variables int pares e impares. Impar con valor 3 e par con valor 2.
- Ahora iterar el ciclo for n veces cada vez calculará el término actual y lo agregará a la suma.
- En cada iteración aumente el valor par e impar con 2.
- Devolver la suma resultante
C++
// C++ program to find sum of N terms of the // series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ... #include <bits/stdc++.h> using namespace std; // Function to return sum of // N term of the series int findSum(int N) { // Initialize the variable int Odd = 3; int Even = 2; int Sum = 0; // Run a loop for N number of times for (int i = 0; i < N; i++) { // Calculate the current term // and add it to the sum Sum += (pow(Odd, 3) - pow(Even, 3)); // Increase the odd and // even with value 2 Odd += 2; Even += 2; } return Sum; } // Driver Code int main() { int N = 10; cout << findSum(N); }
Java
// JAVA program to find sum of N terms of the // series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ... import java.util.*; class GFG { // Function to return sum of // N term of the series public static int findSum(int N) { // Initialize the variable int Odd = 3; int Even = 2; int Sum = 0; // Run a loop for N number of times for (int i = 0; i < N; i++) { // Calculate the current term // and add it to the sum Sum += (Math.pow(Odd, 3) - Math.pow(Even, 3)); // Increase the odd and // even with value 2 Odd += 2; Even += 2; } return Sum; } // Driver Code public static void main(String[] args) { int N = 10; System.out.print(findSum(N)); } } // This code is contributed by Taranpreet
Python3
# Python 3 program for the above approach # Function to calculate the sum # of first N term def findSum(N): # Initialize the variable Odd = 3 Even = 2 Sum = 0 # Run a loop for N number of times for i in range(N): # Calculate the current term # and add it to the sum Sum += (pow(Odd, 3) - pow(Even, 3)) # Increase the odd and # even with value 2 Odd += 2 Even += 2 return Sum # Driver Code if __name__ == "__main__": # Value of N N = 10 # Function call to calculate # sum of the series print(findSum(N)) # This code is contributed by Abhishek Thakur.
C#
// C# program to find sum of N terms of the // series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ... using System; class GFG { // Function to return sum of // N term of the series public static int findSum(int N) { // Initialize the variable int Odd = 3; int Even = 2; int Sum = 0; // Run a loop for N number of times for (int i = 0; i < N; i++) { // Calculate the current term // and add it to the sum Sum += (int)(Math. Pow(Odd, 3) - Math.Pow(Even, 3)); // Increase the odd and // even with value 2 Odd += 2; Even += 2; } return Sum; } // Driver Code public static void Main() { int N = 10; Console.Write(findSum(N)); } } // This code is contributed by Samim Hossain Mondal.
Javascript
<script> // Javascript program to find sum of N terms of the // series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ... // Function to return sum of // N term of the series function findSum(N) { // Initialize the variable let Odd = 3; let Even = 2; let Sum = 0; // Run a loop for N number of times for (let i = 0; i < N; i++) { // Calculate the current term // and add it to the sum Sum += (Math.pow(Odd, 3) - Math.pow(Even, 3)); // Increase the odd and // even with value 2 Odd += 2; Even += 2; } return Sum; } // Driver Code let N = 10; document.write(findSum(N)); // This code is contributed by gfgking. </script>
4960
Complejidad de Tiempo : O(N)
Espacio Auxiliar : O(1), ya que no se ha tomado ningún espacio extra.
Enfoque eficiente :
La secuencia se forma usando el siguiente patrón.
Para cualquier valor N, la forma generalizada de la sucesión dada es:
S N = 4*N 3 + 9*N 2 + 6*N
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find the sum of N terms of the // series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ... #include <bits/stdc++.h> using namespace std; // Function to return sum of // N term of the series int findSum(int N) { return 4 * pow(N, 3) + 9 * pow(N, 2) + 6 * N; } // Driver Code int main() { int N = 10; cout << findSum(N); }
Java
// Java program to find the sum of N terms of the // series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ... import java.util.*; class GFG { // Function to return sum of // N term of the series static int findSum(int N) { return (int) (4 * Math.pow(N, 3) + 9 * Math.pow(N, 2) + 6 * N); } // Driver Code public static void main(String[] args) { int N = 10; System.out.print(findSum(N)); } } // This code is contributed by 29AjayKumar
Python3
# Python 3 program to find the sum of N terms of the # series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ... # Function to calculate the sum # of first N term def findSum(N): return 4 * pow(N, 3) + 9 * pow(N, 2) + 6 * N # Driver Code if __name__ == "__main__": # Value of N N = 10 # Function call to calculate # sum of the series print(findSum(N)) # This code is contributed by Abhishek Thakur.
C#
// C# program to find the sum of N terms of the // series 3^3-2^3, 5^3 - 4^3, 7^3 - 6^3, ... using System; class GFG { // Function to return sum of // N term of the series static int findSum(int N) { return 4 * (int)Math.Pow(N, 3) + 9 * (int)Math.Pow(N, 2) + 6 * N; } // Driver Code public static void Main() { int N = 10; Console.Write(findSum(N)); } } // This code is contributed by ukasp.
Javascript
<script> // JavaScript code for the above approach // Function to return sum of // N term of the series function findSum(N) { return 4 * Math.pow(N, 3) + 9 * Math.pow(N, 2) + 6 * N; } // Driver Code let N = 10; document.write(findSum(N)); // This code is contributed by Potta Lokesh </script>
4960
Complejidad de Tiempo : O(1)
Espacio Auxiliar : O(1)
Publicación traducida automáticamente
Artículo escrito por akashjha2671 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA