Dado un número N, la tarea es encontrar la suma de los cuadrados de signos alternos de los primeros N números naturales, es decir,
1 2 – 2 2 + 3 2 – 4 2 + 5 2 – 6 2 + ….
Ejemplos:
Input: N = 2 Output: 5 Explanation: Required sum = 12 - 22 = -1 Input: N = 8 Output: 36 Explanation: Required sum = 12 - 22 + 32 - 42 + 52 - 62 + 72 - 82 = 36
Enfoque ingenuo: O(N)
El enfoque de fuerza bruta o ingenuo para resolver este problema establece encontrar el cuadrado de cada número del 1 al N y sumarlos con signo alterno para obtener la suma requerida.
- Para cada número en 1 a N, encuentre su cuadrado
- Agregue estos cuadrados con signo alterno
- Esto daría la suma requerida.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find Sum of alternating
// sign Squares of first N natural numbers
#include <iostream>
using namespace std;
// Function to calculate
// the alternating sign sum
int summation(int n)
{
// Variable to store the sum
int sum = 0;
// Loop to iterate each number
// from 1 to N
for (int i = 1; i <= n; i++) {
// The alternating sign is put
// by checking if the number
// is even or odd
if (i % 2 == 1)
// Add the square with the sign
sum += (i * i);
else
// Add the square with the sign
sum -= (i * i);
}
return sum;
}
// Driver code
int main()
{
int N = 2;
cout << summation(N);
return 0;
}
Java
// Java program to find Sum of alternating
// sign Squares of first N natural numbers
class GFG
{
// Function to calculate
// the alternating sign sum
static int summation(int n)
{
// Variable to store the sum
int sum = 0;
// Loop to iterate each number
// from 1 to N
for (int i = 1; i <= n; i++) {
// The alternating sign is put
// by checking if the number
// is even or odd
if (i % 2 == 1)
// Add the square with the sign
sum += (i * i);
else
// Add the square with the sign
sum -= (i * i);
}
return sum;
}
// Driver code
public static void main (String[] args)
{
int N = 2;
System.out.println(summation(N));
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 program to find Sum of alternating # sign Squares of first N natural numbers # Function to calculate # the alternating sign sum def summation(n) : # Variable to store the sum sum = 0; # Loop to iterate each number # from 1 to N for i in range(1, n + 1) : # The alternating sign is put # by checking if the number # is even or odd if (i % 2 == 1) : # Add the square with the sign sum += (i * i); else : # Add the square with the sign sum -= (i * i); return sum; # Driver code if __name__ == "__main__" : N = 2; print(summation(N)); # This code is contributed by AnkitRai01
C#
// C# program to find Sum of alternating
// sign Squares of first N natural numbers
using System;
class GFG
{
// Function to calculate
// the alternating sign sum
static int summation(int n)
{
// Variable to store the sum
int sum = 0;
// Loop to iterate each number
// from 1 to N
for (int i = 1; i <= n; i++) {
// The alternating sign is put
// by checking if the number
// is even or odd
if (i % 2 == 1)
// Add the square with the sign
sum += (i * i);
else
// Add the square with the sign
sum -= (i * i);
}
return sum;
}
// Driver code
public static void Main()
{
int N = 2;
Console.WriteLine(summation(N));
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// JavaScript program to find Sum of alternating
// sign Squares of first N natural numbers
// Function to calculate
// the alternating sign sum
function summation(n)
{
// Variable to store the sum
let sum = 0;
// Loop to iterate each number
// from 1 to N
for (let i = 1; i <= n; i++) {
// The alternating sign is put
// by checking if the number
// is even or odd
if (i % 2 == 1)
// Add the square with the sign
sum += (i * i);
else
// Add the square with the sign
sum -= (i * i);
}
return sum;
}
// Driver code
let N = 2;
document.write(summation(N));
// This code is contributed by Surbhi Tyagi
</script>
Producción:
-3
Enfoque eficiente: O(1)
Existe una fórmula para encontrar la suma de los cuadrados de los primeros n números con signos alternos:
![\[\LARGE 1^{2}-2^{2}+3^{2}-4^{2}+... = (-1)^{n+1} \text{ } \frac{n(n+1)}{2}\]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-c1cb79f89bcaa66b20929ec8fd6c4131_l3.png "Rendered by QuickLaTeX.com")
¿Como funciona esto?
We can prove this formula using induction. We can easily see that the formula is true for n = 1 and n = 2 as sums are 1 and -3 respectively. Let it be true for n = k-1. So sum of k-1 numbers is (-1)k(k - 1) * k / 2 In the following steps, we show that it is true for k assuming that it is true for k-1. Sum of k numbers =(-1)k (Sum of k-1 numbers + k2) =(-1)k+1 ((k - 1) * k / 2 + k2) =(-1)k+1 (k * (k + 1) / 2), which is true.
Por lo tanto, para encontrar la suma de cuadrados de signos alternos de los primeros N números naturales, simplemente calcule la fórmula 
e imprima el resultado.
C++
// C++ program to find Sum of alternating
// sign Squares of first N natural numbers
#include <iostream>
using namespace std;
// Function to calculate
// the alternating sign sum
int summation(int n)
{
// Variable to store the absolute sum
int abs_sum = n * (n + 1) / 2;
// Variable to store the sign
int sign = n + 1 % 2 == 0 ? 1 : -1;
// Variable to store the resultant sum
int result_sum = sign * abs_sum;
return result_sum;
}
// Driver code
int main()
{
int N = 2;
cout << summation(N);
return 0;
}
Java
// Java program to find Sum of alternating
// sign Squares of first N natural numbers
class GFG
{
// Function to calculate
// the alternating sign sum
static int summation(int n)
{
// Variable to store the absolute sum
int abs_sum = n * (n + 1) / 2;
// Variable to store the sign
int sign = n + 1 % 2 == 0 ? 1 : -1;
// Variable to store the resultant sum
int result_sum = sign * abs_sum;
return result_sum;
}
// Driver code
public static void main (String[] args)
{
int N = 2;
System.out.println(summation(N));
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 program to find Sum of alternating # sign Squares of first N natural numbers # Function to calculate # the alternating sign sum def summation(n) : # Variable to store the absolute sum abs_sum = n * (n + 1) // 2; # Variable to store the sign sign = 1 if ((n + 1) % 2 == 0 ) else -1; # Variable to store the resultant sum result_sum = sign * abs_sum; return result_sum; # Driver code if __name__ == "__main__" : N = 2; print(summation(N)); # This code is contributed by AnkitRai01
C#
// C# program to find Sum of alternating
// sign Squares of first N natural numbers
using System;
public class GFG
{
// Function to calculate
// the alternating sign sum
static int summation(int n)
{
// Variable to store the absolute sum
int abs_sum = (int)(n * (n + 1) / 2);
// Variable to store the sign
int sign = n + 1 % 2 == 0 ? 1 : -1;
// Variable to store the resultant sum
int result_sum = sign * abs_sum;
return result_sum;
}
// Driver code
public static void Main()
{
int N = 2;
Console.WriteLine(summation(N));
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// Javascript program to find Sum of alternating
// sign Squares of first N natural numbers
// Function to calculate
// the alternating sign sum
function summation(n)
{
// Variable to store the absolute sum
var abs_sum = n * (n + 1) / 2;
// Variable to store the sign
var sign = n + 1 % 2 == 0 ? 1 : -1;
// Variable to store the resultant sum
var result_sum = sign * abs_sum;
return result_sum;
}
// Driver code
var N = 2;
document.write(summation(N));
// This code is contributed by rutvik_56.
</script>
Producción:
-3