Dado un entero positivo, N . Encuentre la suma del primer N término de la serie-
1 2 , (1 2 +2 2 ), (1 2 +2 2 +3 2 ),….,hasta N términos
Ejemplos :
Entrada : N = 3
Salida : 20
Entrada : N = 1
Salida : 1
Enfoque : La secuencia se forma usando el siguiente patrón. Para cualquier valor N-
Dado 1^2, (1^2+2^2), (1^2+2^2+3^2),….,hasta N términos
![\sum_{i=1}^{i=n} i = \frac{n(n+1)}{2} \\ \sum_{i=1}^{i=n} i^{2}= \frac{n(n+1)(2n+1)}{6} \\ \sum_{i=1}^{i=n} i^{3}= \frac{n^{2}(n+1)^{2}}{4} \\ 1^{2}+(1^{2}+2^{2}....)+(1^{2}+2^{2}....+n^{2})\\ = \sum_{i=1}^{i=n} (1^{2}+2^{2}...+i^{2})\\ =\sum_{i=1}^{i=n}\sum_{i=1}^{i=i}i^{2}\\ =\sum_{i=1}^{i=n} \frac{i(i+1)(2i+1)}{6} \\ =\sum_{i=1}^{i=n} \frac{2i^{3}+3i^{2}+i}{6} \\ =\frac{1}{3}\sum_{i=1}^{i=n} i^{3}+\frac{1}{2}\sum_{i=1}^{i=n} i^{2}+\frac{1}{6}\sum_{i=1}^{i=n} i\\ \text{use above summation formula}\\ =\frac{1}{3}[\frac{n^{2}(n+1)^{2}}{4}]+\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}]+\frac{1}{6}[\frac{n(n+1)}{2}]\\ =\frac{n(n+1)^{2}(n+2)}{12}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-451036b7b3c2c979da5eceffb92c1f96_l3.png "Rendered by QuickLaTeX.com")
S norte = norte * (N+1) 2 * (N+2) / 12
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return sum of
// N term of the series
int findSum(int N){
return N*(N+1)*(N+1)*(N+2)/12;
}
// Driver Code
int main()
{
int N = 3;
cout << findSum(N);
}
Java
// Java program to implement
// the above approach
class GFG {
// Function to return sum of
// N term of the series
static int findSum(int N) {
return N * (N + 1) * (N + 1) * (N + 2) / 12;
}
// Driver Code
public static void main(String args[]) {
int N = 3;
System.out.print(findSum(N));
}
}
// This code is contributed by Saurabh Jaiswal
Python3
# Python 3 program for the above approach # Function to return sum of # N term of the series def findSum(N): return N*(N+1)*(N+1)*(N+2)//12 # Driver Code if __name__ == "__main__": # Value of N N = 3 print(findSum(N)) # This code is contributed by Abhishek Thakur.
C#
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to return sum of
// N term of the series
static int findSum(int N){
return N*(N+1)*(N+1)*(N+2)/12;
}
// Driver Code
public static void Main()
{
int N = 3;
Console.Write(findSum(N));
}
}
// This code is contributed by Samim Hossain Mondal.
Javascript
<script>
// Javascript program to implement
// the above approach
// Function to return sum of
// N term of the series
function findSum(N){
return N*(N+1)*(N+1)*(N+2)/12;
}
// Driver Code
let N = 3;
document.write(findSum(N));
// This code is contributed by Samim Hossain Mondal.
</script>
Producción
20
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