Suma de series con cuadrados alternos con signo de AP

Posted on by Rudeus Greyrat

Nos dan el entero n y también en la siguiente línea 2*n enteros que representan una serie de progresión aritmética a 1 , a 2 , a 3 …a 2n están en AP. Necesitamos encontrar la suma de a 1 2 – a 2 2 + a 3 2 …. + un 2n-1 2 – un 2n 2 .
Ejemplos: 
 

Input : n = 2
        a[] = {1 2 3 4}
Output : -10
Explanation : 12 - 22 + 
32 42 = -10.

Input : n = 3
        a[] = {2 4 6 8 10 12}
Output : -84

Enfoque simple: uno por uno encontramos la suma del cuadrado de la serie con términos pares negativos y el término impar como término positivo. 
 

C++

// CPP program to find sum of
// series with alternate signed
// square AP sums.
#include <bits/stdc++.h>
using namespace std;
 
// function to calculate series sum
int seriesSum(int n, int a[])
{
    int res = 0;
    for (int i = 0; i < 2 * n; i++)
    {
        if (i % 2 == 0)
            res += a[i] * a[i];
        else
            res -= a[i] * a[i];
    }
    return res;
}
 
// Driver Code
int main()
{
    int n = 2;
    int a[] = { 1, 2, 3, 4 };
    cout << seriesSum(n, a);
    return 0;
}

Java

// Java program to find sum of
// series with alternate signed
// square AP sums.
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG
{
 
    // function to calculate
    // series sum
    static int seriesSum(int n,
                         int[] a)
    {
        int res = 0, i;
        for (i = 0; i < 2 * n; i++)
        {
            if (i % 2 == 0)
                res += a[i] * a[i];
            else
                res -= a[i] * a[i];
        }
        return res;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 2;
        int a[] = { 1, 2, 3, 4 };
        System.out.println(seriesSum(n, a));
    }
}

Python3

# Python3 program to find sum
# of series with alternate signed 
# square AP sums.
 
# Function to calculate series sum
def seriesSum(n, a):
    res = 0
     
    for i in range(0, 2 * n):
        if (i % 2 == 0):
            res += a[i] * a[i]
        else:
            res -= a[i] * a[i]
    return res
 
# Driver code
n = 2
a = [1, 2, 3, 4]
print(seriesSum(n, a))
 
# This code is contributed by Ajit.

C#

// C# program to find sum of
// series with alternate signed 
// square AP sums.
using System;
 
class GFG
{
 
    // function to calculate
    // series sum
    static int seriesSum(int n,
                         int[] a)
    {
        int res = 0, i;
        for (i = 0; i < 2 * n; i++)
        {
            if (i % 2 == 0)
                res += a[i] * a[i];
            else
                res -= a[i] * a[i];
        }
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 2;
        int []a = { 1, 2, 3, 4 };
        Console.WriteLine(seriesSum(n, a));
    }
}
 
//This code is contributed by vt_m.

PHP

<?php
// PHP program to find sum of
// series with alternate signed 
// square AP sums.
 
// function to calculate
// series sum
function seriesSum($n, $a)
{
    $res = 0;
    for ( $i = 0; $i < 2 * $n; $i++)
    {
        if ($i % 2 == 0)
            $res += $a[$i] * $a[$i];
        else
            $res -= $a[$i] * $a[$i];
    }
    return $res;
}
 
    // Driver Code
    $n = 2;
    $a = array(1, 2, 3, 4);
    echo seriesSum($n, $a);
 
// This code is contributed by anuj_67.
?>

Javascript

<script>
// JavaScript program to find sum of
// series with alternate signed
 
    // function to calculate
    // series sum
    function seriesSum(n, a)
    {
        let res = 0, i;
        for (i = 0; i < 2 * n; i++)
        {
            if (i % 2 == 0)
                res += a[i] * a[i];
            else
                res -= a[i] * a[i];
        }
        return res;
    }
 
// Driver Code
 
        let n = 2;
        let a = [1, 2, 3, 4];
        document.write(seriesSum(n, a));
 
// This code is contributed by code_hunt.
</script>
Producción : 

-10

 

Complejidad del tiempo: O(2*n)

Espacio auxiliar: O(1) ya que usa espacio constante para variables 

Enfoque eficiente: Uso de la progresión aritmética Aplicación 
Sabemos que la diferencia común d = a 2 – a 1 = a 3 – a 2 = a 4 – a 3
Resultado = a 1 2 – a 2 2 + a 3 2 …. + a 2n-1 2 – a 2n 2
= (a 1 – a 2 )*(a 1 + a 2 ) + (a 3 – a 4 )*(a 3 +a 4 )+….+(a 2n- 1 – un 2n)*(a 2n-1 + a 2n )
Entonces, como la diferencia común es común a la serie, entonces: 
(a 1 – a 2 )[a 1 + a 2 + a 3 …a 2n ]
ahora podemos escribir: 
 

(-d)*(Sum of the term of the 2n term of AP)
(-d)*[((2*n)*(a1 + a2n))/2]
now we know that common difference is : d = (a1 - a2)
Then the difference between : g = (a2n - a1)
So we can conclude that g = d*(2*n - 1)
the we ca replace d by : g/(2*n - 1)

So our result becomes : (n/(2*n - 1)) * (a12 - a2n2)

C++

// Efficient CPP program to
// find sum of series with
// alternate signed square AP sums.
#include <bits/stdc++.h>
using namespace std;
 
// function to calculate
// series sum
int seriesSum(int n, int a[])
{
    return n * (a[0] * a[0] - a[2 * n - 1] *
                a[2 * n - 1]) / (2 * n - 1);
}
 
// Driver code
int main()
{
    int n = 2;
    int a[] = { 1, 2, 3, 4 };
    cout << seriesSum(n, a);
    return 0;
}

Java

// Efficient Java program to 
// find sum of series with
// alternate signed square AP sums.
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG
{
    static int seriesSum(int n,
                         int[] a)
    {
    return n * (a[0] * a[0] - a[2 * n - 1] *
                a[2 * n - 1]) / (2 * n - 1);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int n = 2;
        int a[] = { 1, 2, 3, 4 };
        System.out.println(seriesSum(n, a));
    }
}

Python3

# Efficient Python3 program 
# to find sum of series with 
# alternate signed square AP sums.
 
# Function to calculate
# series sum
def seriesSum(n, a):
 
    return (n * (a[0] * a[0] - a[2 * n - 1] *
                 a[2 * n - 1]) / (2 * n - 1))
 
# Driver code
n = 2
a = [1, 2, 3, 4]
print(int(seriesSum(n, a)))
 
# This code is contributed
# by Smitha Dinesh Semwal.

C#

// Efficient C# program to find sum
// of series with alternate signed
// square AP sums.
using System;
 
class GFG
{
    static int seriesSum(int n, int[] a)
    {
    return n * (a[0] * a[0] - a[2 * n - 1] *
                a[2 * n - 1]) / (2 * n - 1);
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 2;
        int []a= { 1, 2, 3, 4 };
        Console.WriteLine(seriesSum(n, a));
    }
}
 
// This code is contributed by anuj_67..

PHP

<?php
// Efficient PHP program to
// find sum of series with
// alternate signed square AP sums.
 
// function to calculate
// series sum
function seriesSum( $n, $a)
{
    return $n * ($a[0] * $a[0] -
                 $a[2 * $n - 1] *
                 $a[2 * $n - 1]) /
                 (2 * $n - 1);
}
 
    // Driver code
    $n = 2;
    $a = array(1, 2, 3, 4);
    echo seriesSum($n, $a);
     
// This code is contributed by anuj_67..
?>

Javascript

<script>
// Efficient Javascript program to
// find sum of series with
// alternate signed square AP sums.
 
// function to calculate
// series sum
function seriesSum(n, a)
{
    return n * (a[0] * a[0] -
                 a[2 * n - 1] *
                 a[2 * n - 1]) /
                 (2 * n - 1);
}
 
    // Driver code
    let n = 2;
    a = [1, 2, 3, 4];
    document.write(seriesSum(n, a));
     
// This code is contributed by _saurabh_jaiswal.
</script>
Producción : 

-10

 

Complejidad temporal : O(1) desde que se realizan operaciones constantes

Espacio auxiliar: O(1) ya que usa espacio constante para variables 

Publicación traducida automáticamente

Artículo escrito por Manish_100 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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