Encuentre la suma de los primeros n términos de la serie dada:
3, 20, 63, 144, .....
Ejemplos:
Input : n = 2 Output : 23 Input : n =4 Output : 230
Enfoque :
Primero, tenemos que encontrar el término general (Tn) de la serie dada.
series can we written in the following way also: (3 * 1^2), (5 * 2^2), (7 * 3^2), (9 * 4^2), .......up t n terms Tn = (General term of series 3, 5, 7, 9 ....) X (General term of series 1^2, 2^2, 3^2, 4^2 ....) Tn = (3 + (n-1) * 2) X ( n^2 ) Tn = 2*n^3 + n^2
Podemos escribir la suma de la serie de las siguientes maneras:
Sn = 3 + 20 + 63 + 144 + ........up to the n terms Sn = 2 * (sum of n terms of n^3 ) + (sum of n terms of n^2)
Las siguientes son las fórmulas de la suma de n términos de n^3 y n^2:
Below is the implementation of the above approach:
C++
// C++ program to find the sum of n terms #include <bits/stdc++.h> using namespace std; int calculateSum(int n) { return (2 * pow((n * (n + 1) / 2), 2)) + ((n * (n + 1) * (2 * n + 1)) / 6); } int main() { int n = 4; cout << "Sum = " << calculateSum(n) << endl; return 0; }
Java
// Java program to find the sum of n terms import java.io.*; public class GFG { static int calculateSum(int n) { return (int)((2 * Math.pow((n * (n + 1) / 2), 2))) + ((n * (n + 1) * (2 * n + 1)) / 6); } public static void main (String[] args) { int n = 4; System.out.println("Sum = " + calculateSum(n)); } } // This code is contributed by Raj
Python3
# Python3 program to find the sum of n terms def calculateSum(n): return ((2 * (n * (n + 1) / 2)**2) + ((n * (n + 1) * (2 * n + 1)) / 6)) #Driver code n = 4 print("Sum =",calculateSum(n)) # this code is contributed by Shashank_Sharma
C#
// C# program to find the sum of n terms using System; class GFG { static int calculateSum(int n) { return (int)((2 * Math.Pow((n * (n + 1) / 2), 2))) + ((n * (n + 1) * (2 * n + 1)) / 6); } // Driver Code public static void Main () { int n = 4; Console.WriteLine("Sum = " + calculateSum(n)); } } // This code is contributed by anuj_67
PHP
<?php // PHP program to find the // sum of n terms function calculateSum($n) { return (2 * pow(($n * ($n + 1) / 2), 2)) + (($n * ($n + 1) * (2 * $n + 1)) / 6); } // Driver Code $n = 4; echo "Sum = " , calculateSum($n); // This code is contributed by ash264 ?>
Javascript
<script> // javascript program to find the sum of n terms function calculateSum(n) { return parseInt(((2 * Math.pow((n * (n + 1) / 2), 2))) + ((n * (n + 1) * (2 * n + 1)) / 6)); } var n = 4; document.write("Sum = " + calculateSum(n)); // This code contributed by shikhasingrajput </script>
Producción:
Sum = 230
Publicación traducida automáticamente
Artículo escrito por SURENDRA_GANGWAR y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA